## AP 10th Class Telugu Guide Pdf Download – AP Board Solutions Class 10 Telugu

Andhra Pradesh SCERT AP State Board Syllabus SSC 10th Class Telugu Textbook Solutions Study Material Guide Pdf free download are part of AP Board Solutions Class 10.

Students can also read AP 10th Class Telugu Important Questions for board exams.

## AP Board Solutions 10th Class Telugu – AP 10th Class Telugu Study Material Pdf Download

Telugu Guide 10th Class Pdf – AP 10th Class Telugu Textbook Lessons

AP SSC 10th Class Telugu Grammar Question Answers

Telugu 10th Class Guide – AP Board 10th Class Telugu Solutions (Old Syllabus)

## AP 10th Class Maths Chapter 14 Important Questions Probability

These AP 10th Class Maths Chapter Wise Important Questions 14th Lesson Probability will help students prepare well for the exams.

## 14th Lesson Probability Class 10 Important Questions with Solutions

### 10th Class Maths Probability 1 Mark Important Questions

Question 1.
Vineeta said that probability of impossible events is 1. Dhanalakshmi said that probability of sure event is ‘O’ and Sireesha said that probability of any event lies in between 0 and 1. In the above with whom will you agree ?
Sireesha.

Question 2.
A page is opened at a random from a book containing 90 pages. Then find the probability of a page number is a perfect square.
$$\frac{1}{90}$$

Question 3.
From the figure, find the probability of getting blue ball.

$$\frac{3}{5}$$

Question 4.
If P(E) = 0.26, then find P ($$\overline{\mathbf{E}}$$) .
0.74

Question 5.
Which of the following situations have equally likely events ?
Getting 1 or 2 or 3 or 4 or 5 or 6 when a dice is rolled and winning or loosing a game and Head or Tail, when a coin is tossed.

Question 6.
If P(E) is the probability of an event E, then write the relation.
0 ≤ P(E) ≤ 1

Question 7.
The probability of getting right answer to a question is 0.68, then find the
probability of getting a wrong answer.
0.32 (or) 32%

Question 8.
Find the value of P(E) – 1 + p($$\overline{\mathbf{E}}$$).
0

Question 9.
From a well shuffled pack of cards, a card is drawn at random, then find the probability of getting a red jack.
$$\frac{1}{26}$$

Question 10.
There are 50 cards numbered from to 50. A card is drawn at random, then find the probability that the number on the card is divisible by 8.
$$\frac{3}{25}$$

Question 11.
A box contains pencils and pens. The probability of picking out a pen at random is 0.65. Then find the probability of not picking a pen.
0.35

Question 12.
From a bag containing 6 red balls, 5 green balls and 3 blue balls, find the probability of getting a green ball at random.
5/14

Question 13.
If E is an event whose probability is $$\frac{2}{5}$$ then find the probability of not E.
3/5

Question 14.
Three different greeting cards and their corresponding covers are randomly strewn about on a table. If Sita puts the greeting cards into the covers at random, find the probability of correctly matching of all the greeting cards and covers.
$$\frac{1}{6}$$

Question 15.
If a ball is drawn at random from a box containing 11 red balls, 6 white balls and 9 green balls, then find the probability that the ball is not green.
$$\frac{17}{26}$$

Question 16.
From a well shuffled pack of cards a card is drawn at random, then find the probability of getting a red coloured card.
1/2

Question 17.
Find the probability of getting a number less than 5 when a die is rolled.
2/3

Question 18.
If a card is drawn from a deck of 52 cards find the probability that it is a club card.
$$\frac{1}{4}$$

Question 19.
In a single throw of two dice, find the probability of getting a total of 3 or 5.
$$\frac{1}{6}$$

Question 20.
If two events have same chances to happen, then they are called which type of events ?
Equally likely events.

Question 21.
Statement (A) : The probability of win-ning a game is 0.4, then the probability of losing it is 0.6.
Statement (B) : P(E) + P(not E) = 1
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
i) Both A and B are true.

Question 22.
Statement (A) : When two coins are tossed simultaneously, then the probability of getting no tail is $$\frac{1}{4}$$.
Statement (B) : The probability of getting a head (i.e., no tail) in one toss of a coins is $$\frac{1}{2}$$.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
i) Both A and B are true.

Question 23.
Statement (A) : In a simultaneously throw of a pair of dice. The probability of getting a double is $$\frac{1}{6}$$.
Statement (B) : Probability of an event may be negative.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
ii) A is true, B is false

Question 24.
Statement (A) : If P(A) = 0.3 and P(A∪$$\overline{\mathbf{B}}$$) = 0.8, then P(B) is $$\frac{2}{7}$$.
Statement (B) : P ($$\overline{\mathbf{E}}$$) = 1 – P(E), where E is any event.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
i) Both A and B are true.

Question 25.
Statement (A) : If a box contains, 5 white, 2 red and 4 black marbles, then the probability of not drawing a white marble from the box is $$\frac{5}{11}$$.
Statement (B) : p($$\overline{\mathbf{E}}$$) = 1 – P(E), where E is any event.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
(iii) A is false, B is true.

Question 26.
Statement (A) : In rolling a dice, the probability of getting number 8 is zero.
Statement (B) : Its an impossible event.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
(i) Both A and B are true.

Question 27.
Probability
A) 0.95
B) 0.02
C) -0.3

Written description
i) An incorrect assignment
ii) No chance of happening
iii) As much chance of happening as not
iv) Very likely to happen
v) Very little chance of happening
A – (iv), B – (v), C – (i)

Question 28.
Probability
A) 0.5
B) 0

Written description
i) An incorrect assignment
ii) No chance of happening
iii) As much chance of happening as not
iv) Very likely to happen
v) Very little chance of happening
A – (iii), B – (ii)

Question 29.
Written description
A) The probability of a sure event is
B) The probability of impossible event is

Probability
i) 0
ii) 1
iii) 2/7
iv) 12
v) 11/13
vi) 1/12
A – (ii), B – (i)

Question 30.
A card is drawn from a well-shuffled deck of 52 cards randomly. What is the probability of getting a card, which is neither an ace nor a king card ?
$$\frac{11}{13}$$

Question 31.
Write an example for sure event.
When a die is thrown the event of getting a number less than or equal to 6.

Question 32.
What is the probability that 2022 have 53 Sundays ?
1/7

Question 33.
The probability of getting a prime number in a single thrown of dice is _________ .
3/6 or 1/2

Question 34.
If P(E) = 0.3, then P(not E) = __________
(A) 0.3
(B)
(C) 0
(D) 0.7
D) 0.7 .

Question 35.
If P(E) = 0.4, then what is the probability of ‘not E’ ?
Probability of ‘not E’ = 1 – P(E)
= 1 – 0.4 = 0.6

Question 36.
If P(E) = 0.7, then P($$\overline{\mathbf{E}}$$) = ___________
P(E) = 0.7
∴ P($$\overline{\mathbf{E}}$$) = 1 – 0.7 = 0.3

Question 37.
If P(E) = 0.05, what is the probability of not‘E’?
P(not E) = 0.95

Question 38.
If P(E) = 3/4, what is the probability of “not E” ?
Solution:
Probability P(E) = 3/4.
Prabability of “not E” P($$\overline{\mathbf{E}}$$) = 1 – P(E)
= 1 – $$\frac{3}{4}$$ = $$\frac{1}{4}$$

Question 39.
When die is rolled once unbiased what is the probability of getting a multiple of 3 out of possible outcomes ?
Solution:
P(E) = $$\frac{\text { Favourable outcomes }}{\text { Total outcomes }}$$ = $$\frac{2}{6}$$ = $$\frac{1}{3}$$

### 10th Class Maths Probability 2 Mark Important Questions

Question 1.
What is the probability that a number selected from the numbers 1, 2, 3,…, 25 is a prime number, when each of the given numbers is equally likely to be selected ?
Solution:
Given the numbers : 1, 2, 3,…, 25
Toted number of possible outcomes = 25
Prime numbers from 1 to 25 = 2, 3, 5, 7, 11, 13, 17, 19, 23
No. of prime numbers = 9
Let E be the event of probability of getting prime number from 1 to 25.
No. of favourable outcomes = 9
Now P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$
Therefore, P(E) = $$\frac{9}{25}$$.

Question 2.
Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7 ?
Solution:
Given numbers : 1, 2, 3,…, 20
Total numbers = 20
Total no. of possible outcomes = 20
3 multiples from 1 to 20 = 3,6,9,12,15,18
7 multiples from 1 to 20 = 7, 14.
3 or 7 multiples = 3,6, 7,9,12,14,15,18
Total number of 3 or 7 multiples = 8
Let E be the event of probability of getting 3 or 7 multiples from 1 to.20.
No. of favourable outcomes = 8
Now P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$
= $$\frac{8}{20}$$ = $$\frac{2}{5}$$
Therefore, P(E) = $$\frac{2}{5}$$.

Question 3.
Cards marked from the number 2 to 100 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number on the card is a number which is a perfect square.
Solution:
Given the numbers : 2, 3,…, 100
No. of numbers = 99
Total number of possible events = 99
Perfect square numbers from 2 to 100 = 4, 9,16, 25, 36, 49, 64,81, 100
No. of perfect squares = 9
Let E be the probability of getting perfect square.
No. of favourable events = 9
Now P(E) = $$\frac{\text { No. of favourable events }}{\text { Total no. of possible events }}$$
= $$\frac{9}{99}$$ = $$\frac{1}{11}$$
Therefore, P(E) = $$\frac{1}{11}$$.

Question 4.
An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
Solution:
Given urn contains 10 red and 8 white balls.
Total no. of balls 10 + 8 = 18
Total no. of possible outcomes = 18
Let E be the event of probability of getting white balls.
No. of white balls = 8
No. of favourable outcomes = 8
Now P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$
= $$\frac{8}{18}$$ = $$\frac{4}{9}$$
Therefore, P(E) = $$\frac{4}{9}$$.

Question 5.
In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize ?
Solution:
Given 10 prizes and 25 blanks
Total no. of card = 10 + 25 = 35
Total no. of possible outcomes = 35
Let E be the event of probability of getting prize.
No. of prizes = 10
No. of favourable outcomes = 10
Now, P(E) = Now P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$
= $$\frac{10}{35}$$ = $$\frac{2}{7}$$
Therefore, P(E) = $$\frac{2}{7}$$.

Question 6.
If the probability of winning a game is 0.3. What is the probability of loosing it ?
Solution:
Given the probability of winning a game is 0.3
That is P(winning) = 0.3
P(winning) + P(loosing) = 1
0.3 + P(loosing) = 1
P(loosing) = 1 – 0.1
Therefore, probability of loosing game = 0.9

Question 7.
The probability of an event is always in between 0 and 1. Why ?
Solution:
From the definition of probability P(E), the numerator (number of outcomes favourable to the event E) is always less than of equal to the denominator (the number of all possible outcomes).
So, the probability of an event is always in between 0 and 1.
i.e., 0 ≤ P(E) ≤ 1.

Question 8.
Find the probability of getting a sum of the numbers on them is 7, when two dice are rolled at a time.
Solution:
When two dice are rolled at a time the total outcomes are = 62 = 36
Number of outcomes such that their sum of numbers on face is 7 = 6
∴ Probability of getting sum of numbers on faces to be 7 = $$\frac{6}{36}$$ = $$\frac{1}{6}$$

Question 9.
Find the probability of getting a prime number, when a card drawn at random from the numbered cards from 1 to 25.
Solution:
Favourable outcomes of prime numbers from 1 to 25 = 9
Total number of outcomes = 25
Probability of getting a prime number
= $$\frac{\text { Number of favourable outcomes }}{\text { Total number of total outcomes }}= [latex]\frac{9}{25}$$ Question 10. From the first 50 natural numbers, find the probability of randomly selected number is a multiple of 3. Solution: Multiples of 3 from 1 to 50 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48 No. of favourable outcomes to get multiples of 3 from 1 to 50 = 16 Probability of getting multiple of 3 from 1 to 50 = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$ = $$\frac{16}{50}$$ = $$\frac{8}{25}$$. Question 11. A dice is thrown once. Find the probability of getting a composite number. Solution: Possible outcomes = 1, 2, 3, 4, 5, 6 Favourable outcomes = 4, 6 Probability of a getting a compositive number = $$\frac{\text { Number of favourable outcomes }}{\text { Total possible outcomes }}$$ = $$\frac{2}{6}$$ = $$\frac{1}{3}$$. Question 12. What is the probability of getting exactly two heads, when three coins tossed simultaneously ? Solution: Probability of getting exactly two heads = $$\frac{\text { number of favourable out comes }}{\text { total number of out comes }}$$ = $$\frac{3}{8}$$ Question 13. From English alphabet if a letter is choosen at random, then find the probability that the letter is a consonant. Solution: Number of total outcomes = 26 Number of favourable outcomes = 21 Probability that the letter is consonant = $$\frac{\text { Number of favourable outcomes }}{\text { Number of total outcomes }}$$ = $$\frac{21}{26}$$ Question 14. If a dice is rolled once, then find the probability of getting an odd number. Solution: Total outcomes = {1, 2, 3, 4, 5, 6} Number of Total outcomes = 6 Favourable outcomes = {1, 3, 5} Number of favourable outcomes = 3 P (an odd number) = $$\frac{\text { Number of favourable outcomes }}{\text { Number of Total outcomes }}$$ = $$\frac{3}{6}$$ = $$\frac{1}{2}$$ Question 15. Find the probability of getting a ‘vowel’ if a letter is chosen randomly from the word “INNOVATION”. Solution: P(E) = $$\frac{\text { Number of favourable outcomes }}{\text { Number of total outcomes }}$$ = $$\frac{3}{6}$$ = $$\frac{1}{2}$$ Favourable outcomes = i, o, a, i, o Number of favourable outcomes = 5 Number of total outcomes = 10 P(getting a vowel) = $$\frac{5}{10}$$ = $$\frac{1}{2}$$ Question 16. You are writing a test of 40 objective type questions. Each question carries 1 mark. What is the probability of marks you may get to be in multiple of 5 ? Solution: Total Number of questions = 40 Total Number of outcomes = 40 Number of Multiples of 5 upto 40 = 8 Favourable outcomes = 8 Probability for getting multiples of 5 = \begin{aligned}&\text { Favourable outcomes for getting }\\&\frac{\text { multiples of } 5}{\text { Total No. of possible outcomes }} \end{aligned} = $$\frac{8}{40}$$ = $$\frac{1}{5}$$ Question 17. A page is opened at random from a book containing 100 pages. Find the probability that the page number is a perfect square. Solution: Number of pages in given book = 100 The page numbers that will be perfectly a square number (If randomly that are selected) are 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. ∴ Number of favourable outcomes = 100 Number of all possible outcomes = its number of pages = 100 ∴ Probability of getting a perfect number = $$\frac{10}{100}$$ = 0.1. Question 18. A box contains 3 blue and 4 red balls. What is the probability that the ball taken out randomly will be red ? Solution: Total number of balls in the box = 3 + 4 = 7 No. of favourable outcomes for picking a red ball = 4 ∴ Probability of red ball = $$\frac{\text { Number of favourable outcomes }}{\text { Total outcomes }}$$ ∴ P(E) = $$\frac{4}{7}$$ Question 19. A three digit number is formed by the digits 2, 3 and 5 without repetition. What is the probability that the number is divisible by 5 ? Let ’E’ be the event of choosing a three digit number divisible by 5. All possible three digit numbers (without repetition) 235, 253, 325, 352, 523, 532. ∴ n(S) = 6; E = {235, 325}; n(E) = 2 ∴ p(E) = $$\frac{2}{6}$$ = $$\frac{1}{3}$$ Question 20. In a classroom, 32 students out of 60 can take tea. Find the probability of “The tea not taken”. Solution: Total Number of possible outcomes = 60 No. of students doesn’t take tea (No. of favourable outcomes) = 60 – 32 = 28 Probability of students not taken tea = $$\frac{\text { No, of favourable outcomes for not taken tea }}{\text { Total No. of possible outcomes }}$$ = $$\frac{28}{60}$$ = $$\frac{7}{15}$$ Question 21. If P(E) = 0.25, what is the probability of ‘not E’ ? Solution: Given P(E) = 0.25 Hence P(E) + P($$\overline{\mathbf{E}}$$) = 1 Where P($$\overline{\mathbf{E}}$$) is the probability of ‘not E’. ⇒ 0.25 + P($$\overline{\mathbf{E}}$$) = 1 ⇒ P($$\overline{\mathbf{E}}$$) = 1 – 0.25 = 0.75. Question 22. Find the probability of getting the letter M in the word. “MATHEMATICS”. Solution: Let the event of getting the letter M be E Number of favourable outcomes to E = 2 Number of all possible outcomes = 11 ∴ P(E) = $$\frac{\text { No. of favourable outcomes to E }}{\text { All possible outcomes }}$$ = = $$\frac{2}{11}$$ ### 10th Class Maths Probability 4 Mark Important Questions Question 1. Find the probability that a leap year selected at random will contain 53 Sundays. Solution: In every year there are 365 days. But in a leap year there are 366 days. 366 days = 52 weeks + 2 days. So, every leap year 52 Sundays and the remaining 2 days can be possible in 7 events. They are, 1) Sunday and Monday 2) Monday and Tuesday 3) Tuesday and Wednesday 4) Wednesday and Thursday 5) Thursday and Friday 6) Friday and Saturday 7) Saturday and Sunday. So, total no. of possible outcomes = 7 Let E be the event of probability of getting 53 Sundays. No. of favourable outcomes = 2 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ ∴ P(E) = $$\frac{2}{7}$$ Question 2. It is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb ? Solution: Given no. of bulbs in the box = 600 Total no. of bulbs = 600 Total no. of possible outcomes = 600 Let E be the event of probability of getting a non-defective bulb. No. of defective bulbs = 12 No. of non-defective bulbs = 600 – 12 = 588 No. of favourable outcomes = 588 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ = $$\frac{588}{600}$$ = $$\frac{49}{50}$$ ∴ P(E) = 0.98. Question 3. A letter is chosen at random from the letters of the word ASSASSINATION. Find the probability that the letter chosen is a i) Vowel ii) Consonant. Solution: In the given word “ASSASSINATION” there are 13 letters. So, total no. of possible outcomes = No. of letters = 13 i) No. of vowels in the word = 6 No. of favourable outcomes = 6 Let E be the event of probability of getting vowel in the given word. Now,P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ = $$\frac{6}{13}$$ ii) No. of consonants in the word = 7 Let E be the event of probability of getting consonant in the given word. No. of favourable outcomes = 7 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ P(E) = $$\frac{7}{13}$$ Question 4. A jar contains 24 marbles some are green and other are blue. If a marble is drawn at random from the jar, the probability that it is green is $$\frac{2}{3}$$ . Find the number of blue marbles in the jar. Solution: Given the no. of marbles in the jar = 24 Let the no. of green marbles = x The no. of blue marbles = 24 – x Total no. of possible outcomes = 24 Let E be the event of probability of getting green marble. No. of favourable outcomes of getting green marbles = x Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ Given P(E) = $$\frac{x}{24}$$ = $$\frac{2}{3}$$ x = $$\frac{2}{3}$$ × 24 = 16 No. of green marbles (x) = 16 No. of blue marbles = 24 – x = 24 – 16 = 8 Therefore, no. of blue marbles = 8. Question 5. A number of x is selected from the numbers 1, 2, 3 and then a second number 7 is randomly selected from the numbers 1, 4, 9. What is the probability that the product xy of the two numbers will be less than 9 ? Solution: Given a number x is selected from 1, 2, 3 and y is selected from 1, 4, 9 Therefore, two numbers can be selected in 9 ways as given below. (x, y) as (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (2, 9), (3, 1), (3, 4), (3, 9) So, total no. of possible outcomes = 9. Let E be the event of probability of getting the product xy of the two numbers will be less than 9. Favourable outcomes = (1, 1), (1, 4), (2, 1), (2, 4), (3, 1) No. of favourable outcomes = 5 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ ∴ P(E) = $$\frac{5}{9}$$. Question 6. What is the probability that a number selected from the numbers 1, 2, 3,…, 15 is a multiple of 4 ? Solution: Given numbers : 1, 2, 3,…, 15 Total numbers = 15 Total no. of possible outcomes = 15 4 multiples from 1 to 15 are 4, 8, 12 No. of 4 multiples = 3 No. of favourable outcomes = 3 Let E be the event of probability of getting 4 multiples Now, P(E) $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ = $$\frac{3}{15}$$ = $$\frac{1}{5}$$ ∴ P(E) = $$\frac{1}{5}$$. Question 7. A number is selected at random from the first 50 natural numbers. Find the probability it is a multiple of 3 and 4. Solution: Given numbers : 1, 2, 3,…, 50 Toted numbers = 50 Total no. of possible outcomes = 50 Multiples of 3 and 4 from 1 to 50 = 12, 24, 36,48 Let E be the event of probability of getting a number of multiple of 3 and 4. No. of multiples of 3 and 4 = 4 No. of favourable outcomes = 4 Now, P(E) $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ = $$\frac{4}{50}$$ = $$\frac{2}{25}$$ ∴ P(E) = $$\frac{2}{25}$$. Question 8. There are 12 red, 18 blue and 6 white balls in a box. When balls is drawn at random from the box, what is the probability of not getting a red ball ? Solution: Total Number of balls = 12 + 8 + 6 = 36 Number of Red balls = 12 ∴ Probability of getting Red ball P(R) = $$\frac{\text { favourable outcomes }}{\text { total outcomes }}$$ = $$\frac{12}{36}$$ = $$\frac{1}{3}$$ ∴ Probability of not getting Red ball p($$\overline{\mathrm{R}}$$)= 1 – $$\frac{1}{3}$$ = $$\frac{2}{3}$$ (Or) Total Number of balls = 12 + 18 + 6 = 36 Exclude, the Red balls, the number of remaining balls = 18 + 6 = 24 Probability of not getting a Red ball = $$\frac{24}{36}$$ = $$\frac{2}{3}$$ Question 9. When a card is drawn from a well shuffled deck of 52 cards, then find the probability of NOT getting a red faced card. Solution: Number of total outcomes = 52 Number of red face cards = 6 Probability of a red face card = $$\frac{6}{52}$$. ∴ Probability of “NOT” getting a red face card = 1 – $$\frac{6}{52}$$ = $$\frac{52 – 6}{52}$$ = $$\frac{46}{52}$$ = $$\frac{23}{26}$$ Question 10. There are 5 red balls, 4 green balls and 6 yellow balls in a box. If a ball is selected random, what is the probability of not getting a yellow ball ? Solution: Total number of balls in a bag = 15 Total number of chance to select a ball from a bag = 15 Favourable outcomes to select not yellow ball = 9 Probability of not getting a yellow ball = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$$ = $$\frac{9}{15}$$ = $$\frac{3}{5}$$. Question 11. One card is selected from a well – shuffled deck of 52 cards. Find the probability of getting a red card with prime number. Solution: Number of cards in a deck = 52 Number of red card with prime number = 8 Probability of getting red card with prime number P(E) = $$\frac{\text { Number of favourable outcomes }}{\text { Number of total outcomes }}$$ = $$\frac{8}{52}$$ = $$\frac{2}{13}$$. Question 12. From the following data, find the prob-ability of selecting ‘B’ blood group student.  Blood group A B AB O Number of students 10 13 12 5 Solution: Number of favourable outcomes = 13 Number of total possible outcomes = 40 ∴ Probability of selecting B blood group = $$\frac{\text { No.of Favourable outcomes }}{\text { No. of Total outcomes }}$$ = $$\frac{13}{40}$$ Question 13. What is the probability of a number picked from first twenty natural numbers is even composite number ? Solution: Sample space = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …………….. 20} Favourable out comes = {4, 6, 8, 10, 12, 14, 16, 18, 20} Probability of getting even composite number from first twenty natural numbers = $$\frac{\text { Number of favourable outcomes }}{\text { Number of total outcomes }}$$ = $$\frac{9}{20}$$ Question 14. A bag contains balls which are numbered from 1 to 50. A ball is drawn at random from the bag, the probability that it bears a two digit number multiple of 7. Solution: Number of possible outcomes = 50 Number of required outcomes = 6 {14, 21, 28, 35, 42, 49} \ Probability of getting two digit number which is a multiple of 7 = $$\frac{\text { Number of favourable outcomes }}{\text { Number of total outcomes }}$$ = $$\frac{6}{50}$$ = $$\frac{3}{25}$$ Question 15. A box contains 4 red balls, 5 green balls and P white balls. If the probability of randomly picked ball from the box to be a red ball is $$\frac{1}{3}$$, then find the number of white balls. Solution: P(E) = $$\frac{\text { No. of outcomes }}{\text { Sample Space }}$$ = $$\frac{1}{3}$$ = $$\frac{4}{4+5+P}$$ 4 + 5 + P = 12 ⇒ P = 3 ∴ Number of white balls = 3 Question 16. A bag contains 7 red, 5 white and 6 black balls. A ball is drawn from the bag at random, find the probability that the ball drawn is not black. Solution: Number of total outcomes = 7 + 5 + 6 = 18 Number of favourable outcomes = 7 + 5 = 12 Probability that the ball dravn is not black = $$\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }}$$ = $$\frac{12}{18}$$ = $$\frac{2}{3}$$ Question 17. A box contains 20 cards which are numbered from 1 to 20. If one card is selected at random from the box, find the probability that it bears (i) a prime number, (ii) an even number. Solution: Total possibilities = {1, 2, 3, 4, 5, 6, 7, 8, 9, …….., 20} Number of Total possibilities = 20 i) Prime numbers = {2, 3, 5, 7, 11, 13, 17, 19} Number of favourable outcomes = 8 P(Prime Number) = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$$ = $$\frac{8}{20}$$ = $$\frac{2}{5}$$ ∴ P(Prime Number) = $$\frac{2}{5}$$ ii) Even numbers = {2, 4, 6, 8,10, 12, 14,16,18, 20} Number of favourable outcomes = 10 P (Even Number) = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$$ = $$\frac{10}{20}$$ = $$\frac{1}{2}$$ ∴ P(Even number) = $$\frac{1}{2}$$ Question 18. A box contains four slips numbered 1, 2, 3, 4 and another box contains five slips numbered 5, 6, 7, 8, 9. If one slip is taken randomly from each box, i) How many number pairs are possible ? ii) What is the probability of both being odd ? iii) What is the probability of getting the sum of the numbers 10 ? Solution: i) Possible number pairs (1, 5) (1, 6) (1, 7) (1, 8) (1, 9) (2, 5) (2, 6) (2, 7) (2, 8) (2, 9) (3, 5) (3, 6) (3, 7) (3, 8) (3, 9) (4, 5) (4, 6) (4, 7) (4, 8) (4, 9) Number of possible number pairs = 20 ii) Favourable outcomes (1, 5), (1, 7), (1, 9), (3, 5), (3, 7), (3, 9) Number of favourable outcomes = 6 P(getting both being odd) = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$$ = $$\frac{6}{20}$$ = $$\frac{3}{10}$$ iii) Favourable outcomes (1, 9), (2, 8), (3, 7), (4, 6) Number of favourable outcomes = 4 P(getting the sum of the numbers 10) = $$\frac{4}{20}$$ = $$\frac{1}{5}$$ Question 19. In a bag, there are 5 Red balls, 2 Black balls and 3 White balls. If one ball is selected randomly from the bag, then find the probability of – i) getting a Red ball. ii) getting not a Red ball. Solution: Red balls = 5 Black balls= 2 White balls = 3 Total balls = 5 + 2 + 3 = 10 Number of favourable i) P(E) = $$\frac{\text { outcomes }}{\text { Total number of outcomes }}$$ P (getting a Red ball) = $$\frac{5}{10}$$ = $$\frac{1}{2}$$ ii) P (Not getting a Red ball) = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$$ = $$\frac{2 + 3}{10}$$ = $$\frac{5}{10}$$ = $$\frac{1}{2}$$ Question 20. There are 100 flash cards labelled from 1 to 100 in a bag. When a card is drawn from the bag at random, what is the probability of getting ……… i) a card with prime number from possible outcomes ? ii) a card without prime number from possible outcomes ? Answer: Number of prime numbers between 1 and 100 = 25 i) Probability of getting a card with prime numbers = $$\frac{25}{100}$$ = $$\frac{1}{4}$$ = 0.25 ii) Probability of getting a card without prime number = $$\frac{75}{100}$$ = 0.75 Question 21. A shopkeeper has 100 memory cards in a box. Among them, 15 memory cards are defective. When a person came to the shop to buy a memory card, the shopkeeper drew a memory card at random from the box. Then, i) what is the probability that this memory card is defective ? ii) after drawing the first memory card which is defective, it is not placed back in the box. Then another memory card is drawn at random. What is the probability that this memory card is NOT defective ? Solution: i) Total all possible outcomes = n(T) = 100 Number of favourable outcomes = n(E) = 15 The probability that memory card is defective P(E) = $$\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{T})}$$ = $$\frac{15}{100}$$ = $$\frac{3}{20}$$ ii) Removed defective card is not placed in the box, so the probability that memory card is defective = P(E’) = $$\frac{14}{100}$$ = $$\frac{7}{50}$$ The probability that memory card is NOT defective = 1 – P(E’) = 1 – $$\frac{7}{50}$$ = $$\frac{50 – 7}{50}$$ = $$\frac{43}{50}$$ Question 22. A die is thrown once. Find the probability of getting (i) a prime number (ii) a number lying between 1 and 5. Solution: When a die is thrown once sample space = S = {1, 2, 3, 4, 5, 6} Total number of outcomes = n(S) = 6 i) Let ‘E’ be an event of getting a prime number. Favourable outcomes to E = {2, 3, 5} n(E) = 3 ∴ P(E) = $$\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{3}{6}$$ = $$\frac{1}{2}$$ ii) Let ‘F’ be an event of getting a number lying between 1 and 5. Favourable outcomes to F = {2, 3, 4} n(F) = 3 ∴ P(F) = $$\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{3}{6}$$ = $$\frac{1}{2}$$ ### 10th Class Maths Probability 8 Mark Important Questions Question 1. Two unbiased coins are tossed simultaneously. Find the probability of getting i) two heads ii) one head iii) one tail iv) atleast one head v) at most one head vi) no head Solution: If two unbiased coins are tossed simultaneously. We get the outcomes HH, HT, TH, TT. Total number of possible outcomes = 4. i) Let E be the event of probability of getting two heads. So, favourable outcomes = HH. No. of favourable outcomes = 1 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ P(E) = $$\frac{1}{4}$$ ii) Let E be the event of probability of getting one head. So, favourable outcomes = HT, TH. No. of favourable outcomes = 2 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ P(E) = $$\frac{2}{4}$$ = $$\frac{1}{2}$$ ∴ P(E) = $$\frac{1}{2}$$ iii) Let E be the event of probability of getting one tail. So, favourable outcomes = TH, HT. No. of favourable outcomes = 2 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ P(E) = $$\frac{2}{4}$$ = $$\frac{1}{2}$$ ∴ P(E) = $$\frac{1}{2}$$ iv) Let E be the event of probability of getting atleast one head. So, favourable outcomes = HT, TH, HH No. of favourable outcomes = 3 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ ∴ P(E) = $$\frac{3}{4}$$ v) Let E be the event of probability of getting at most one head. So, favourable outcomes . = HT, TH, HH No. of favourable outcomes = 3 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ ∴ P(E) = $$\frac{3}{4}$$ vi) Let E be the event of probability of getting no head. So, favourable outcomes = TT No. of favourable outcomes = 1 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ ∴ P(E) = $$\frac{1}{4}$$ Question 2. Find the probability that a number selected at random from the numbers 1, 2, 3,.., 35 is a i) prime number ii) multiple of 7 iii) a multiple of 3 or 5 Solution: Given numbers are 1, 2, 3,.., 35 Total number of possible outcomes = 35 i) Let E be the event of probability of getting prime number. Prime numbers from 1 to 35 = 2, 3, 5, 7,11,13,17, 19, 23, 29,31 No. of favourable outcomes = 11 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ ∴ P(E) = $$\frac{11}{35}$$ ii) Let E be the event of probability of getting 7 multiple from 1 to 35. 7 multiples : 7, 14, 21, 28, 35 No. of favourable outcomes = 5 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ P(E) = $$\frac{5}{35}$$ = $$\frac{1}{7}$$ ∴ P(E) = $$\frac{1}{7}$$ iii) Let E be the event of probability of getting multiple of 3 or 5. 3 multiples = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 5 multiples = 5, 10, 15, 20, 25. 30, 35 Favourable outcomes = 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35 No. of favourable outcomes = 16 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ ∴ P(E) = $$\frac{16}{35}$$. Question 3. Cards marked with numbers 13, 14, 15 … 60 are placed in a box and mixed thoroughly one card is drawn at random from the box. Find the probability that number on the card drawn is i) divisible by 5 ii) a number is a perfect square. Solution: Given number cards = 13, 14, 15,…, 60 Total number of cards = 48 Total number of possible outcomes = 48 i) 5 multiples from 13 to 60 are: 15,20, 25, 30, 35, 40, 45, 50, 55, 60. Let E be the event of probability of getting of 5 multiples. No. of favourable outcomes = 10 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ P(E) = $$\frac{10}{48}$$ = $$\frac{5}{24}$$ ∴ P(E) = $$\frac{5}{24}$$ ii) Let E be the event of probability of getting a perfect square. Perfect squares from 13 to 60 are 16, 25, 36, 49. No. of perfect squares = 4 No. of favourable outcomes = 4 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ P(E) = $$\frac{4}{48}$$ = $$\frac{1}{12}$$ ∴ P(E) = $$\frac{1}{12}$$. Question 4. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probabili¬ty of getting a card of i) heart ii) queen iii) clubs Solution: Given king, queen and jack of clubs removed from deck. Then number of cards remaining = 52 – 3 = 49 Total number of possible outcomes = 49 i) Let E be the event of probability of getting heart. Number of cards of heart =13 No. of favourable outcomes = 49 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ ∴ P(E) = $$\frac{13}{49}$$. ii) Let E be the event of probability of getting queen. No. of queens in the deck = 4 But, one queen clubs removed. So, number of queens = 4 – 1 = 3 No. of favourable outcomes = 3 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ ∴ P(E) = $$\frac{3}{49}$$. iii) Let E be the event of probability of getting clubs. No. of cards of clubs in the deck = 13 But, 3 cards (king, queen and jack) of clubs removed. So, no. of cards of club = 13 – 3 = 10 No. of favourable outcomes = 10 Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ ∴ P(E) = $$\frac{10}{49}$$. Question 5. A target shown in the given figure consists of three concentric circles of radii 3, 7 and 9 cm respectively. A dart is thrown and lands on the target. What is the probability that the dart will land on the shaded region ? Solution: Given radii of concentric circles 3 cm, 7 cm and 9 cm Total area of the figure = πr2 = π × 92 = 81π. sq. cm Total possible area = 81π. sq. cm Area of the circle of radius 3 cm = π × 32 = 9π. sq. cm Area of the circle of radius 7 cm = π × 72 = 49π. sq. cm Area of the shaded region = 49π – 9π = (49 – 9) π = 40π. sq. cm Let E be the probability of the dart land on the shaded region. Favourable area = 40π. sq. cm Now, P(E) = $$\frac{\text { No. of favourable area }}{\text { Total no. of possible area }}$$ = $$\frac{40 \pi}{81 \pi}$$ ∴ P(E) = $$\frac{40}{81}$$. Question 6. A square dart board is shown in the figure. The length of a side of the larger square is 1.5 times the length of a side of the smaller square. If a dart is thrown and lands on the larger square. What is the probability that it will interior of the smaller square ? Solution: Given ABCD and PQRS are two squares. Let side of square (PQRS) = x units. Area of square PQRS = x.x = x2 sq. units. Then side of the square ABCD = 1.5 times of x = 1.5 × x Area of square ABCD = 1.5x × 1.5x = 2.25x2 sq. units. Total possible area = 2.25x2 sq. units. Favourable area of a dart get into the smaller square = x2 sq. units. Let E be the event of probability of landing the dart into smaller square. Now, P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$$ P(E) = $$\frac{1 x^2}{2.25 x^2}$$ = $$\frac{1 \times 100}{2.25 \times 100}$$ = $$\frac{100}{225}$$ = $$\frac{4}{9}$$ ∴ P(E) = $$\frac{4}{9}$$. Question 7. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of red ball, find the number of blue balls in the bag. Solution: Number of red balls present in a bag = 5 Let the No.of blue balls = x (say) Then the total No.of balls = 5 + x From those (5 + x) balls in the bag the number of favourable outcomes to take a red ball randomly = 5 So the probability of taking a red ball = $$\frac{5}{5+x}$$ Now The number of favourable outcomes to take a blue ball randomly = x So the probability of taking a blue ball = $$\frac{x}{5+x}$$ From the given problem Probability of blue bell = (Probability of red ball) (2) $$\frac{x}{5+x}$$ = [latex]\frac{5}{5+x} × 2
∴ $$\frac{x}{5+x}$$ = $$\frac{10}{5+x}$$ ⇒ x = 10
∴ No. of blue balls in the bag = 10

Question 8.
From the deck of 52 cards, if a card is randomly chosen, find the probability of getting a card with (i) a prime number on it, (ii) face on it.
Solution:
The number cards in a deck = 52
Total number of outcomes = 52
i) Cards with a prime number on it = {2, 3, 5, 7} (in one suit)
Number of favourable outcomes = 4 × 4 = 16 (Q In each suit there are 4 cards with primes) Probability of getting a card with a prime number on it = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$$
= $$\frac{16}{52}$$ = $$\frac{4}{13}$$

ii) Number of face cards = 12
Number of favourable outcomes = 12
Probability of getting a face
card = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$$
= $$\frac{12}{52}$$ = $$\frac{3}{13}$$

Question 9.
Suppose you drop a dice at random on the circular region of diameter 28 cm as shown in the figure. What is the probability that it will land inside the rectangle ?
Solution:

Diameter of circle = d = 28 cm
r = d/2 = 14 cm
Area of the circle = πr2
= $$\frac{22}{7}$$ × 14 × 14
= 22 × 28 = 616 cm2
Area of the rectangle = l × b = 11 × 7 = 77 cm2
∴ Probability that it will land inside the rectangle = $$\frac{\text { Area of the Rectangle }}{\text { Area of the circle }}$$
= $$\frac{77}{616}$$ = $$\frac{7}{56}$$ = $$\frac{1}{8}$$

Question 10.
A bag contains 20 discs, which are numbered from 1 to 20. If one disc is drawn at random from the bag, find the probability that it bears:
i) an even number,
ii) Prime number,
iii) Multiple of 5,
iv) Two digit odd number.
Solution:
Total number of possible outcomes = 20
i) For probability of the disc bears an even number
No. of favourable outcomes = 10 Probability = $$\frac{\text { No. of favourable outcomes }}{\text { Total No. of possible outcomes }}$$ = $$\frac{10}{20}$$ = $$\frac{1}{2}$$

ii) For probability of the disc bears a prime number
No.of favourable outcomes = 8
Probability = $$\frac{8}{20}$$ = $$\frac{2}{5}$$

iii) For probability of the disc bears a multiple of 5
No. of favourable outcomes = 4
Probability = $$\frac{4}{20}$$ = $$\frac{1}{5}$$

iv) For probability of the disc bears a two digit odd number
No. of favourable outcomes = 5
Probability = $$\frac{5}{20}$$ = $$\frac{1}{4}$$

Question 11.
Two dice are thrown at the same time. What is the probability that the sum of two numbers appearing on the top of the dice is (a) 10, (b) less than or equal to 12, (c) a prime number, (d) multiple of ‘3’ ?
Solution:
Total number of possible outcomes when rolling two dice at a time = 6 × 6 = 36
Favourable outcomes of getting each sum is 10.
a) Sum be 10 = {(5, 5), (4, 6), (6, 4)}
No. of favourable outcomes = 3
∴ Required probability = P(E) = $$\frac{3}{36}$$ = $$\frac{1}{12}$$

(b) The outcomes favourable to the event
“Less than or equal to 12” be denotes by ‘F’ are
= {(1, 1) (1,2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
No. of outcomes favourable to ‘F’ is n(F) = 36
∴ P(F) = $$\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{36}{36}$$ = 1

c) The outcomes favourable to the event “Sum of two numbers a prime number” be denoted by ‘G’ are (1, 1) (1, 2) (1, 4) (1, 6) (2,1) (2, 3) (2, 5) (3, 2) (3, 4) (4, 1) (4, 3) (5, 2) (5, 6) (6, 1) (6, 5)
No. of outcomes favourable to ‘G’ is n(G) = 15
∴ P(G) = $$\frac{\mathrm{n}(\mathrm{G})}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{15}{36}$$ = $$\frac{5}{12}$$

d) The outcomes favourable to the event “Sum of two numbers be multiple of 3” be denoted by ‘H’,are (1, 2) (1, 5) (2, 1) (2, 4) (3, 3) (3, 6) (4, 2) (4, 5) (5, 1) (5, 4) (6, 3) (6, 6)
No. of outcomes favourable to ‘H’ is n(H) = 12
∴ P(H) = $$\frac{\mathrm{n}(\mathrm{H})}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{12}{36}$$ = $$\frac{1}{3}$$

Question 12.
Two digit numbers are formed by the digits 0, 1, 2, 3, 4, where the digits are not repeated. Find the probability that
i) the number formed is greater than 42.
ii) the number formed is a multiple of 4.
Solution:
Two digit numbers formed by the digits 0, 1, 2, 3, 4
where the digits are not repeated
(10, 12, 13, 14, 20, 21, 23, 24, 30, 31, 32, 34, 40, 41, 42, 43)
∴ Sample space = (10, 12, 13, 4, 20, 21, 23, 24, 30, 31, 32, 34, 40, 41, 42, 43)
∴ n(S) = 16
i) Probability of getting the number formed is greater than 42
= $$\frac{\text { No. of possible outcomes }}{\text { Total outcomes }}$$
here no. of possible outcomes = 1 (that is 43 only)
∴ Probability = $$\frac{1}{16}$$ ______ (1)

ii) In the sample space multiples of ‘4’ = 12, 20, 24, 32, 40
∴ No. of multiples of 4 = 5
Now probability for forming a multiple of ’4’ = $$\frac{5}{16}$$ _____ (2)

Question 13.
A box contains 1 to 100 number cards. If one card is drawn at random. Find the probability of that card will be
i) a perfect square
ii) a prime number
iii) a two digit number
iv) a multiple of 9.
Solution:
Total number of possible outcomes = n(S) = 100
i) Let ‘E’ be an event of getting a perfect square.
Favourable outcomes to E = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
n(E) = 10.
∴ P(E) = $$\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{10}{100}$$ = $$\frac{1}{10}$$

ii) Let ‘F’ be an event of getting a prime number.
Favourable outcomes to
F = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
n(F) = 25. ∴ P(F) = $$\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{25}{100}$$ = $$\frac{1}{4}$$

iii) Let ‘G’ be an event of getting a two digit number.
Favourable outcomes to G = {10 to 99}
n(G) = 90. ∴ P(G) = $$\frac{\mathrm{n}(\mathrm{G})}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{90}{100}$$ = $$\frac{9}{10}$$

iv) Let ‘H’ be an event of getting a multiple of 9.
Favourable outcomes to
H = {9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99}
n(H) = 11.
∴ P(H) = $$\frac{\mathrm{n}(\mathrm{H})}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{11}{100}$$

Question 14.
A dice is thrown twice. What is the probability that
i) 3 will come up at least once ?
ii) 3 will not come up either time ?
Solution:
When a dice is thrown once sample space = S = (1, 2, 3, 4, 5, 6}
Total number of outcomes = n(S) = 6
A dice thrown twice, then total number of outcomes = n(S) = 6 × 6 = 36

i) Let ‘E’ be an event of getting 3 will come up at least once.
Favourable outcomes to E = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3), (6, 3)}
n(E) = 11
∴ P(E) = $$\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{11}{36}$$

ii) Probability that 3 will not come up either time = F = 36 – 11 = 25

Question 15.
One card is drawn from a well – shuffled deck of 52 cards. Find the probability of getting
(i) a king of black colour
(ii) a face card
(iv) a card not a heart
Solution:
When one card is drawn from a well shuffled deck of 52 cards.
The total number of outcomes in sample space is n(S) = 52

Probability = $$\frac{\text { No.of favourable outcomes }}{\text { Total No.of outcomes }}$$
Event i) : Consider E1 be the Event of getting a king of black colour.
Number of favourable outcomes for the event E1 is n(E1) = 2

P(E1) = $$\frac{\mathrm{n}\left(\mathrm{E}_1\right)}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{2}{52}$$ = $$\frac{1}{26}$$.

Event ii) : Consider E2 be the Event of getting a face card.
Number of favourable outcomes for the event E2 is n(E2) = 12
P(E2) = $$\frac{\mathrm{n}\left(\mathrm{E}_2\right)}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{12}{52}$$ = $$\frac{3}{13}$$.

Event iii) : Consider E3 be the Event of getting a spade.
Number of favourable outcomes for the event E3 is n(E3) = 13
P(E3) = $$\frac{\mathrm{n}\left(\mathrm{E}_3\right)}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{13}{52}$$ = $$\frac{1}{4}$$.

Event iv) : Consider E4 be the Event of getting a card not to be a heart.
Number of favourable outcomes for the event E4 is n(E4) = 39
P(E4) = $$\frac{\mathrm{n}\left(\mathrm{E}_4\right)}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{39}{52}$$ = $$\frac{3}{4}$$.

Question 16.
One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:
i) A queen of black colour
ii) a face card
iii) a jack of diamond
iv) a club card
Solution:
Total number of cards in a deck n(S) = 52
i) Probability of getting “A queen of black colour”.
Number of favourable outcomes for getting a queen of black colour.
n(Q) = 2
∴ P(Q) = $$\frac{\text { Number of favourable outcomes }}{\text { Total Number of all possible outcomes }}$$
∴ P(Q) = $$\frac{2}{52}$$ = $$\frac{1}{26}$$

ii) Probability of getting “a face card”. Number of favourable for getting face card = 12
∴ P(F) = $$\frac{12}{52}$$ = $$\frac{3}{13}$$

iii) Probability of getting “a jack of diamond”.
Number of favourable outcomes for getting a jack of diamond = 1
∴ P(J) = $$\frac{1}{52}$$

iv) Probability of getting a club card. Number of favourable outcomes for getting a club card = 13
∴ P(C) = $$\frac{13}{52}$$ = $$\frac{1}{4}$$

Question 17.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
i) A face card of diamond
ii) Ace card
iv) A jack of red
Solution:
Total number of cards in a deck = 52.
∴ Number of all possible outcomes n(S) = 52
i) Number of outcomes favourable to the face card of diamond n(D) = 3
∴ Probability of getting face card of diamond
P(D) = $$\frac{\text { No.of favourable outcomes }}{\text { No.of total outcomes }}$$
= $$\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{3}{52}$$

ii) Number of outcomes favourable to ace card n(A) = 4
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}$$ = $$\frac{4}{52}$$ = $$\frac{1}{13}$$

iii) Number of outcomes favourable to spade card n(SP) = 13
∴ P(SP) = $$\frac{13}{52}$$ = $$\frac{1}{4}$$

iv) Number of outcomes favourable to jack of red n(JR) = 2
∴ P(JR) = $$\frac{2}{52}$$ = $$\frac{1}{26}$$

## AP 10th Class Maths Chapter 13 Important Questions Statistics

These AP 10th Class Maths Chapter Wise Important Questions 13th Lesson Statistics will help students prepare well for the exams.

## 13th Lesson Statistics Class 10 Important Questions with Solutions

### 10th Class Maths Statistics 1 Mark Important Questions

Question 1.
If the mean and the median of a data are 12 and 15 respectively, then find its mode.
Solution:
Mode = 3 median – 2 mean
= 3 × 15 – 2 × 12 = 45 – 24 = 21.

Question 2.
If every term of the statistical data consisting of n terms is decreased by 2, then the mean of the data :
A) decreases by 2
B) remains unchanged
C) decreases by 2n
D) decreases by 1
Solution:
A) decreases by 2

Question 3.
For the following distribution :

Find the lower limit of modal class.
Solution:
Lower limit of modal class = $$\frac{15 + 15}{2}$$ = $$\frac{30}{2}$$ = 15

Question 4.
Consider the following frequency distribution :

Find the median class.
Solution:
N = 12 + 10 + 15 + 8 + 11 = 56
$$\frac{N}{2}$$ = $$\frac{56}{2}$$ = 28

$$\frac{N}{2}$$ = $$\frac{56}{2}$$ = 28
.’. Median classes = 12 – 18.

Question 5.
For the following distribution :

The sum of lower limits of median class and modal class.
Solution:
Modal Class = 15 – 20

 Class frequency cf 0 – 5 10 10 5 – 10 15 25 10 – 15 12 37 15 – 20 20 57 20 – 25 9 66 N = 66

$$\frac{N}{2}$$ = $$\frac{66}{2}$$ = 33
Median Class = 10 – 15
lower limit of median class + lower limit of model class 10 + 15 = 25

Question 6.
Find the median class for the data given below.

Solution:

 Class frequency cf 20 – 40 10 10 40 – 60 12 22 60 – 80 14 36 80 – 100 13 49 100 – 120 17 66 N = 66

$$\frac{N}{2}$$ = $$\frac{66}{2}$$ = 33
Median Class = 60 – 80

Question 7.
Mean and median of some data are 32 and 30 respectively. Using empirical relation, find the mode of the data.
Solution:
Mode = 3 Median – 2 mean
= 3 × 30 – 2 × 32 = 90 – 64 = 26

Question 8.
If the mean of the first n natural number is 15, then find n.
Solution:
Mean of first ‘n’ natural numbers
= $$\frac{\mathrm{n}(\mathrm{n}+1)}{2 \mathrm{n}}$$ = $$\frac{(n+1)}{2}$$ = $$\frac{(n+1)}{2}$$ = 15
= n + 1 = 30 = n = 30 – 1 = n = 29.

Question 9.
Find the class marks of the classes 20 – 50 and 35 – 60.
Solution:
Class marks of 20 – 50 = $$\frac{20 + 50}{2}$$ = $$\frac{70}{2}$$
= 35
Class mark of 35 – 60 = $$\frac{35 + 60}{2}$$ = $$\frac{95}{2}$$ = 47.5

Question 10.
For the following frequency distribution.

Find the upper limit of median class.
Solution:

 Class Frequency cf 0-5 8 8 5 – 10 10 18 10 – 15 19 37 15 – 20 25 62 20 – 25 8 70 N = 20

$$\frac{N}{2}$$ = $$\frac{70}{2}$$
Median class = 10 – 15
Upper limit = 15

Question 11.
If the mean of the following distribution is 2.6, then find the value of y.

Solution:

 x f fx 1 4 4 2 5 10 3 y 3y 4 1 4 5 2 10

N = 12 + y Σfx = 28 + 3y
Mean = 2.6
$$\frac{28+3 y}{12+y}$$ = 2.6 ⇒ 28 + 3y = 2.6(12 + y)
28 + 3y = 31.2 + 2.6y
3y – 2.6y = 31.2 – 28
0.4y = 3.2 ⇒ y = $$\frac{32}{4}$$ ⇒ y = 8

Question 12.
The time in seconds, taken by 150 athletes to turn a 100 m hurdle race are tabulated below:

Find the number of athletes who completed the race in less than 17 seconds.
Solution:

 Class Frequency cf 13 – 14 2 2 14 – 15 4 6 15 – 16 5 11 16 – 17 71 82 17 – 18 48 130 18 – 19 20 150 N = 150

$$\frac{N}{2}$$ = $$\frac{150}{2}$$
Number of athlets who completed the race lessthan 17 records is 82.

Question 13.
Find the median of first seven prime numbers.
Solution:
The first seven prime numbers are 2, 3, 5, 7, 11, 13, 17
median = 7

Question 14.
If the mean of 6, 7, x, 8, y, 14 is 9, then
A) x + y = 21
B) x + y = 19
C) x – y = 19
D) x – y = 21
Solution:
B) x + y = 19
Mean = $$\frac{6+7+x+8+y+14}{6}$$
9 = $$\frac{35+y+x}{6}$$
54 = 35 + x + y
x + y = 54 – 35
x + y = 19

Question 15.
Find the mode of the numbers 2, 3, 3, 4, 5, 4, 4, 5, 3, 4, 2, 6, 7.
Solution:
4 repeated more times
∴ Mode = 4

Question 16.
Write the empirical relationship between the three measures of central tendency.
Solution:
Mode = 3 median – 2 mean (or)
2 Mean = 3 median – mode (or)
3 Median = Mode + 2 mean

Question 17.
Median and Mode of a distribution are 25 and 21 respectively. Find the mean of the data using empirical relationship.
Solution:
Mode = 3 median – 2 mean
21 = 3 x 25 – 2 × mean
21 = 75 – 2 mean
2 mean = 75 – 21
2 mean = 54
Mean = $$\frac{54}{2}$$ = 27

Question 18.
Find the class-marks of the classes 10 – 25 and 35 – 55.
Solution:
Class marks of 10 – 25 = $$\frac{35}{2}$$ = 17.5
Class marks of 35 – 55 = $$\frac{90}{2}$$ = 45

Question 19.
Write the mean of first ’n’ natural numbers.
Solution:
Mean = $$\frac{1+2+3+\ldots \ldots n}{n}$$
= $$\frac{\mathrm{n}(\mathrm{n}+1)}{2 \mathrm{n}}$$ = $$\frac{\mathrm{n}+1}{2}$$

Question 20.
If xi and fi are numerically small, then which method is appropriate choice to find mean.
Solution:
Direct method

Question 21.
_________ based on all observations.
A) Range
B) Mean
C) Median
D) Mode
Solution:
A) Range

Question 22.
If the median of a series exceeds the mean by 3. Find by what number the mode exceeds its mean.
Solution:
Given, median – mean = 3
median = 3 + mean
mode = 3 median – 2 mean
mode = 3(3 + mean) – 2 mean
mode = 9 + 3 mean – 2 mean
mode = 9 + mean
mode – mean = 9

Question 23.
Find the median class from the following frequency distribution.

Solution:
N = 8 + 15 + 21 + 8 = 52
$$\frac{N}{2}$$ = $$\frac{52}{2}$$

Median class = 1550 – 1700

Question 24.
The median of the data, using an emperical relation when it is given that mode = 12.4, Mean = 10.5
Solution:
Mode = 3 median – 2 mean
12.4 = 3 median -2 (10.5) ⇒ 12.4 = 3 median – 21
12.4 + 21 = 3 median ⇒ 33.4 = 3 median
Median = $$\frac{33.4}{3}$$ = 11.13

Question 25.
In the following frequency distribution, the median class.

Solution:

$$\frac{N}{2}$$ = $$\frac{100}{2}$$ = 50
Median Class = 150 – 155

Question 26.
Mode of a data is 10, one observation 4 is added to the data then find the mode.
Solution:
There is no change in mode.
∴ Mode = 10

Question 27.
Find the mode of letters A, B, C, D ….Z.
Solution:
No mode

Question 28.
Mean of a data is 11. Each observation is added by 2 then find the mean of new observation.
Solution:
New mean = 11 + 2 = 13

Question 29.
Assertion (A) : Am of 7, 4, 9 is 8.2
Reason (R) : Am = $$\frac{\text { Sum of observations }}{\text { No. of observations }}$$
A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true.
Solution:
D) Assertion (A) is false but Reason (R) is true.

Question 30.
Assertion (A) : Midvalue of class 0 – 10 is 5
Reason (R) : Range = Maximum value – Minimum value
A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true.
Solution:
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Question 31.
Assertion (A) : Mode of 1, 2, 3, … 10 is 10.
Reason (R) : Mode = 3 median – 2 mean
A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason- (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true.
Solution:
D) Assertion (A) is false but Reason (R) is true.

Question 32.
Assertion (A) : The mean of the following data is 412.

 x 1 2 3 4 5 f 1 2 3 4 5

Reason (R) : Mean, $$\bar{x}$$ = $$\frac{\sum f_i x_i}{\sum f_i}$$.
A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true.
Solution:
D) Assertion (A) is false but Reason (R) is true.

Question 33.
Assertion (A) : Mean of x, y, z is 4 then mean of 2x, 2y, 2z is 8.
Reason (R) : If there are two values in a data after arranging either in ascen-ding or descending order, the average of those two values is called median of that data.
A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true.
Solution:
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Question 34.
Assertion (A) Midvalues are used to calculate mean.
Reason (R) In a grouped data, Mean = l + ($$\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}$$) × h
A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true.
Solution:
C) Assertion (A) is true but Reason (R) is false.

Question 35.
Assertion (A) : In step-deviation method of finding mean, ui = $$\frac{x_i-a}{h}$$.
Reason (R) : A data may or may not have mode.
A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertioh (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true. .
Solution:
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Question 36.
In the formula of mean of grouped data in step deviation method $$\bar{x}$$ = A + ($$\frac{\sum f_i u_i}{\sum f_i}$$) × h; where ui = ________
Solution:
ui = $$\frac{x_i-a}{h}$$.

Question 37.
In the classes 35 – 39, 40 – 44, 45 – 49,…. of a frequency distribution, then the upper boundary of the class 40 – 44 is __________ .
Solution:
44.5

Question 38.
Assertion (A) : Mode of sin 0°, cos 0°, sin 90° and tan 45° is 0.
Reason (R) : $$\bar{x}$$ = $$\frac{\sum f_{f_1}}{\sum f_i}$$
A) Both Assertion and Reason are true. Reason is supporting the Assertion.
B) Both Assertion and Reason are true. But Reason is not supporting the Assertion.
C) Assertion is true but the Reason is false.
D) Assertion is false but the Reason is true.
Solution:
D) Assertion is false but the Reason is true.

Question 39.
From the given frequency distribution table, what is the class interval of highest frequency class ?

Solution:
2 (class 3 – 5)

Question 40.
Find the mean of the given data :
2, 3, 7, 6, 6, 3, 8
Solution:
Data = 2, 3, 7, 6, 6, 3, 8
Numbers = 7
Total = 2 + 3+7 + 6 + 6 + 3 + 8 = 35
Mean = $$\frac{35}{7}$$ = 5

### 10th Class Maths Statistics 2 Mark Important Questions

Question 1.
Find the mean the following distribution :

 Class Frequency 3 – 5 5 5 – 7 10 7 – 9 10 9 – 11 7 11 – 13 8

Solution:

Mean = $$\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}$$ = $$\frac{326}{40}$$ = $$\frac{32.6}{4}$$ = 8.15

Question 2.
Find the mode of the following data :

 Class Frequency 0 – 20 6 20 – 40 8 40 – 60 10 60 – 80 12 80 – 100 6 100 – 120 5 120 – 140 3

Solution:

l = 60 ; f1 = 12 ; f0 = 10 ; f2 = 6 h = 20
Mode = l + ($$\frac{f_1-f_0}{2 f_1-f_0-f_2}$$) × h
= 60 + ($$\frac{12-10}{2 \times 12-10-6}$$) × 20
= 60 + $$\frac{2 \times 20}{24-16}$$
= 60 + $$\frac{40}{8}$$
= 60 + 5 = 65.

Question 3.
Find the mode of the following frequency distribution :

Solution:

l = 30 ; f1 = 10 ; f0 = 9 ; f2 = 3 ; h = 5
Mode Z = l + ($$\frac{f_1-f_0}{2 f_1-f_0-f_2}$$) × h
= 30 + ($$\frac{10-9}{2 \times 10-9-3}$$) × 5
= 30 + ($$\frac{5}{20-12}$$) = 30 + $$\frac{40}{8}$$
Mode = 30 + 0.625 = 30.625

Question 4.
Find the mean for the following distribution :

Solution:

Mean, $$\overline{\mathbf{x}}$$ = $$\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}$$
= $$\frac{230}{10}$$ = 23

Question 5.
The following distribution shows the transport expenditure of 100 employees :

Solution:
Find the mode of the distribution.

l = 400 ; f0 = 25 ; f1 = 21 ; f2 = 19 ; h = 200
Mode, Z = l + ($$\frac{f_1-f_0}{2 f_1-f_0-f_2}$$) × h
= 400 + ($$\frac{25-21}{2 \times 25-21-19}$$) × 200
= 400 + ($$\frac{4 \times 200}{50-40}$$) × 400 + $$\frac{800}{10}$$
= 400 + 80 = 480

Question 6.

How do you find the deviation from the assumed mean for the above data ?
Solution:
The assumed value in calculation of mean of a grouped data is the mid value of the class interval which has maximum frequency.

Question 7.
When an observation in a data is abnormally more than or less than the remaining observations in the data, does it affect the mean or mode or median ? Why ?
Solution:
Mean and medians are depending upon given data except mode. So when an observa¬tion in a data is abnormally changes it affects those values i.e., mean and median.

Question 8.
Write the formula to find the mean of a grouped data, using assumed mean method and explain each term.
Solution:
Mean = a + $$\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}$$
a – assumed mean, f – frequency,
d – x – a, x – class mark

Question 9.
The median of observations, -2, 5, 3, -1, 4, 6 is 3.5″. Is it correct ? Justify your answer.
Solution:
Yes.
Given data : -2, 5, 3, -1, 4, 6
Ascending order : -2, -1, 3, 4, 5, 6
Median = $$\frac{3 + 4}{2}$$ = $$\frac{7}{2}$$ = 3.5

Question 10.
Prathyusha stated that “the average of first 10 odd numbers is also 10”. Do you agree with her ? Justify your answer.
Solution:
The average of first 10 odd numbers
= $$\frac{10}{2}$$ $$\frac{[1+19]}{10}$$ = $$\frac{5 \times 20}{10}$$ = $$\frac{100}{10}$$ = 10
∴ The average of first 10 odd numbers is 10.
I agree with Prathyusha statement. l

Question 11.
Write the formula to find the median of a grouped data and explain the alphabet in it.
Solution:
Median = l + $$\left[\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right]$$ × h
l = lower boundary of median class,
n =number of observations.
cf = cumulative frequency of class preceeding the median class,
f = frequency of median class,
h = class size.

Question 12.
Find the median of first seven composite numbers.
Solution:
The first seven composite numbers are

∴ Median = 9

Question 13.
Find the mode of the data 6, 8, 3, 6, 3, 7, 4, 6, 7, 3, 6.
Solution:
First we arrange the ascending order
3, 3, 3, 4, 6, 6, 6, 6, 7, 7, 8

 x frequency 3 3 4 1 6 4 7 2 8 1

∴ Mode (6) [having maximum frequency]

Question 14.
Find the mean of prime numbers which are less than 30.
Solution:
Mean = $$\frac{\text { sum of the observation }}{\text { number of the observation }}$$
= $$\frac{2+3+5+7+11+13+17+19+23+29}{10}$$ = $$\frac{129}{10}$$ = 12.9

Question 15.
Find the median of $$\frac{2}{3}$$, $$\frac{4}{5}$$, $$\frac{1}{2}$$, $$\frac{3}{4}$$, $$\frac{6}{5}$$, $$\frac{1}{2}$$.
Solution:

∴ Median = $$\frac{3}{4}$$

### 10th Class Maths Statistics 4 Mark Important Questions

Question 1.
ArIthmetic mean of the following data is 14. FInd the value of k.

 xi 5 10 15 20 25 fi 7 k 8 4 5

Solution:

Mean = $$\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$$ = $$\frac{360+10 \mathrm{k}}{24+\mathrm{k}}$$ = 14
360+ 10k = 14(24 + k)
360 + 10k = 336 + 14k
10k – 14k = 336 – 360
-4k = -24
k = $$\frac{24}{4}$$ = 6
∴ k = 6

Question 2.
If the mean of the following data is 18.75. Find the value of p.

 xi 10 15 15 25 30 fi 5 10 10 8 2

Solution:

Mean = $$\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$$ = $$\frac{460+7 p}{32}$$ = 18.75
460 + 7p = 18.75 × 32
460 + 7p = 600
7p = 600 – 460
7p = 140
P = $$\frac{140}{7}$$ = 20
∴ p = 20

Question 3.
The heights of six members of a family are given below in the table.

 Height in feet 5 5.2 5.4 5.6 No. of family members 1 2 2 1

Find the mean height of the family members.
Solution:

 Height in feet (x) No. of family members (f) f × x 5 1 5 5.2 2 10.4 5.4 2 10.8 5.6 1 5.6 Σf = 6 Σf × x = 31.8

∴ Mean height of the family members
= $$\frac{\Sigma \mathrm{fx}}{\Sigma \mathrm{f}}$$ = $$\frac{31.8}{6}$$ = 5.3

Question 4.
State the formula to find the mode for a grouped data. Explain each term In it
(OR)
Write the formula of mode for grouped data and explain each term in it.
Solution:
Formula to find the mode for a grouped data
Mode = Z = l + ($$\frac{\Sigma \mathrm{fx}}{\Sigma \mathrm{f}}$$) × h
Where
l = Lower boundary of the modal class
h = Size of the modal class interval
f1 = Frequency of the modal class
f0 = Frequency of the class preceding the modal class
f2 = Frequency of the class succeeding the modal class

Question 5.

Find the value of Σfixi for the above data, where xi is the mid value of each class.
Solution:

Question 6.
The height of 12 members of a family are given below in the table.

 Height (in ft) 5 5.2 5.4 5.6 No.of family members 3 4 3 2

Find the mean height of the family members.
Solution:

 Height (x) No.of family Members (f) f.x 5 3 15 5.2 4 20.8 5.4 3 16.2 5.6 2 11.2 Σf = 12 Σfx = 63.2

Mean height of the family = $$\frac{\Sigma \mathrm{fx}}{\Sigma \mathrm{f}}$$
= $$\frac{63.2}{12}$$ = 5.266 feet = 5.27 feet

Question 7.
Find the median of first six prime numbers.
Solution:
First six prime numbers

Median of the above data,
If n is even then the median will be the average of the ($$\frac{n}{2}$$)th and ($$\frac{n}{2}$$ + 1)th observations.
($$\frac{n}{2}$$)th term = ($$\frac{6}{2}$$) = 3rd term
($$\frac{n}{2}$$ + 1)th term = ($$\frac{6}{2}$$ + 1)th = 3 + 1 = 4th term
Median = ($$\frac{3^{\text {rd }} \text { term }+4^{\text {th }} \text { term }}{2}$$) = $$\frac{5 + 7}{2}$$ = $$\frac{6}{2}$$ = 6
∴ Median of first six prime number is 6.

### 10th Class Maths Statistics 8 Mark Important Questions

Question 1.
Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

 xi 10 30 50 70 90 fi 17 f1 32 f2 19

N = 120
Solution:

68 + f1 + f2 = 120
f1 + f2 = 120 – 68
f1 + f2 = 52 → (1)
Mean = $$\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$$ = $$\frac{3480+30 f_1+70 f_2}{120}$$ = 50
3480 + 30f1 + 70f2 = 50 × 120
30f1 + 70f2 = 6000 – 3480
30f1 + 70f2 = 2520
3f1 + 7f2 = 252 → (2)
(1) × 3 3f1 + 3f2 = 3 × 52
3f1 + 3f2 → (3)

Put f2 = 24 in (1)
f1 + 24 = 52 ⇒ f1 = 52 – 24 ⇒ f1 = 28
∴ f1 = 28, f2 = 24

Question 2.
If the mean of the following distributions is 54, find the value of p.

Solution:

Mean = $$\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$$ = $$\frac{2370+30 p}{39+p}$$ = 54
2370 + 30p = 54 (39 + p)
2370 + 30p = 2106 + 54p
2370 – 2106 = 54p – 30p
264 = 24p
P = $$\frac{264}{24}$$ = 11
∴ P = 11

Question 3.
The following table gives weekly wages in rupees of workers in a certain commer¬cial organization. The frequency of class 43 – 46 is missing. It is known that the mean of the frequency distribution is 47.2. Find the missing frequency.

Solution:

Mean = $$\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$$ = 47.2
$$\frac{7803+44.5 \mathrm{x}}{162+\mathrm{x}}$$ = 47.2
7803 + 44.5x = 47.2 (162 + x)
7803 + 44.5x = 47.2 x 162 + 47.2x
7803 – 7646.4 = 47.2x – 44.5x
156.6 = 2.7x
x = $$\frac{156.6}{2.7}$$ = 58
∴ x = 58

Question 4.
The mean of the following frequency table 50. But the frequencies f1 and f2 in class 20 – 40 and 60 – 80 are missing. Find the missing frequencies.

Solution:

Note : Highest frequency class can be treated as the mean class.
Σfi = 68 + f1 + f2 = 120
f1 + f2 = 120 – 68
f1 + f2 = 52 → (1)
Mean = a + ($$\frac{\sum f_i u_i}{\sum f_i}$$) × h
a = class mark of the assumed mean class = 50
Σfi = Total frequency = 120
h = height of the class
Σfiui = -f1 + f2 + 4

Put f2 = 24 in (1)
f1 + 24 = 52
f1 = 52 – 24 = 28 ⇒ ∴ f1 = 28, f2 = 24

Question 5.
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f1 and f2.

Solution:

Note : Highest frequency class can be treated as the mean class.
Σfi = 30 + f1 + f2 = 50
f1 + f2 = 50 – 30 = 20
f1 + f2 = 20 → (1)
Mean = a + ($$\frac{\sum f_i u_i}{\sum f_i}$$) × h
a = class mark of the assumed mean class = 50
xi = class mark
Σfi = Total frequency = 50
h = height of the class = 20
Σfiui = -f1 + f2 + 28

-f1 + f2 + 28 = 32
-f1 + f2 = 32 – 28
-f1 + f2 = 4 → (2)

Put f2 = 12 in (1)
f1 + 12 = 20
f1 = 20 – 12 = 8
∴ f1 = 8, f2 = 12

Question 6.
If the median of the following frequency distribution is 46. Find the missing frequencies. Total frequency is 229.

Solution:

150 + a + b = 229
a + b = 229 – 150
a + b = 79 → (1)
Median = l + ($$\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}$$) × h
l = lower limit of the median class = 40
n = Total frequency = 229
f = frequency of the median class = 65
h = height of the class = 10
cf = cumulative frequency of the class preceding = 42 + a
Median = 40 + ($$\frac{\frac{229}{2}-(42+\mathrm{a})}{65}$$) × 10 = 46 ⇒ ($$\frac{\frac{229}{2}-(42-\mathrm{a})}{65}$$) × 10 = 46 – 40
⇒ ($$\frac{229}{2}$$ – $$\frac{84}{2}$$ – 2a) 10 = 6 × 65

a = 34 (approximately)
Put a = 34 in (1)
a + b = 79
34 + b = 79 ⇒ b = 79 – 34 = 45 ⇒ ∴ a = 34 and b = 45

Question 7.
Compute the median of the following data

 Marks Number of students More than 150 0 More than 140 12 More than 130 27 More than 120 60 More than 110 105 More than 100 124 More than 90 141 More than 80 150

Solution:

Note : Highest frequency class can be treated as the median class.
Median = l + ($$\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}$$) × h
l = lower limit of the median class = 110, n = Total frequency = 150,
f = frequency of the median class = 45, h = height of the class = 10
cf = cumulative frequency of the class preceding = 45 O50
Median = 110 + ($$\frac{\frac{150}{2}-45}{45}$$) × 10 = 110 + ($$\frac{75 – 45 }{45}$$) × 10 = 110 + $$\frac{30 \times 10}{45}$$
= 110 + 6.66 ⇒ ∴ Median = 116.66

## AP 10th Class Maths Chapter 12 Important Questions Surface Areas and Volumes

These AP 10th Class Maths Chapter Wise Important Questions 12th Lesson Surface Areas and Volumes will help students prepare well for the exams.

## 12th Lesson Surface Areas and Volumes Class 10 Important Questions with Solutions

### 10th Class Maths Surface Areas and Volumes 1 Mark Important Questions

Question 1.
If the radii of two spheres are in the ratio 2 : 3, then find the ratio of their respective volumes.
Solution:
r1 : r2 = 2 : 3
V1 : V2 = $$\frac{2}{3}$$πr13 : $$\frac{2}{3}$$r23
r13 : r23
23 : 33
8 : 27
Given that, Area of base (πr2) = 156
Vertical height of cone (h) = 8 cm

Question 2.
Find the volume of a right circular cone whose area of the base is 156 cm2 and the vertical height is 8 cm.
Solution:
V = $$\frac{1}{3}$$πr2h = $$\frac{1}{3}$$ × 156 × 8
= 52 × 8
= 416 cm2

Question 3.
Find the total surface area of a solid hemisphere of radius 7 cm.
Solution:
r = 7 cm
TSA of hemisphere = 3πr2
= 3 × $$\frac{22}{7}$$ × 7 × 7 cm2
= 66 × 7 cm2
= 462 cm2
(or) TSA of hemisphere = 3πr2
= 3 × π × 7 × 7
= 147π cm2

Question 4.
The curved surface area of a right circular cylinder of height 14cm is 88 cm2. Find the diameter of its circular base.
Solution:
Cylinder, h = 14 cm
CSA = 88 cm2
2πrh = 88
2 × $$\frac{22}{7}$$ × r × 14 = 88
44 × r × 2 = 88
r = 1
2r = 2 × 1
r = 1
d = 2r
d = 2 cm

Question 5.
A solid is of the form of a cone of radius V surmounted on a hemisphere of the same radius. If the height of the cone Is the same as the diameter of Its base, then find the volume of the solid.

Solution:
Volume of solid = Vcone + Vhemisphere
= $$\frac{1}{3}$$πr2h + $$\frac{2}{3}$$πr3
= $$\frac{1}{3}$$πr2[h + 2r]
= $$\frac{1}{3}$$πr2(r + 2r)(∴ h = r)
= $$\frac{1}{3}$$πr2 × 3r
= πr3

Question 6.
Assertion (A) : The surface area of largest sphere that can be inscribed in a hollow cube of side ‘a’ cm is πa2 cm2.
Reason (R) : The surface area of a sphere of radius ‘r’ is $$\frac{4}{3}$$πr3.

A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of the Assertion (A).
C) Assertion (A) is true, but Reason (R) is false.
D) Assertion (A) is false, but Reason (R) is true.
Solution:
C) Assertion (A) is true, but Reason (R) is false.
A : side of cube = a
radii of sphere = $$\frac{\mathrm{a}}{2}$$
surface area = 4πr2
= 4π($$\frac{\mathrm{a}}{2}$$)2
= 4π$$\frac{a^2}{4}$$
= πa2
A is true.
R : Surface area of sphere = 4 × πr2 R is wrong.

Question 7.
The curved surface area of a cylinder is 264 m2 and its volume is 924 m2. Then find the ratio of height to its diameter.
Solution:
Cylinder : 2πrh = 264
πr2h = 924
$$\frac{\pi r^2 h}{2 \pi r h}$$ = $$\frac{924}{264}$$
$$\frac{r}{2}$$ = $$\frac{924}{264}$$
r = $$\frac{924}{132}$$
r = 7
d = 2 × 7 = 14 cm
2 × $$\frac{22}{7}$$ × 7 × h = 264
h = $$\frac{264}{44}$$
h = 6
h : d = 6 : 14 = 3 : 7

Question 8.
A rectangular sheet of paper 40 cm × 22 cm is rolled to form a hollow cylinder of height 40 tin then find the radius of cylinder.
Solution:
Rectangle: P = 2(l + b)
P = 2(40 + 22)
P = 2(62)
P = 124 m
Cylinder, h = 40 cm
2πrh = 124
2 × $$\frac{22}{7}$$ × r × 40 = 124
r × 40 = 124 × $$\frac{7}{22}$$ × $$\frac{1}{2}$$
r = 62 × $$\frac{7}{22}$$ × $$\frac{1}{40}$$
r = $$\frac{31 \times 7}{11 \times 40}$$ = $$\frac{217}{440}$$

Question 9.
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere ?
Solution:
$$\frac{2}{3}$$πr3 =2πr2
$$\frac{r}{3}$$ = 1
r = 3
d = 2 × 3
d = 6

Question 10.
Diagonal of a cuboid is ………….
$$\sqrt{l^2+\mathrm{b}^2+\mathrm{h}^2}$$

Question 11.
Volumes of two spheres are in the ratio 64 : 27 find the ratio of their surface areas.
Solution:
$$\frac{4}{3}$$πr13 : $$\frac{4}{3}$$23 = 64 : 27
r13 : r23 = 43 : 33
r1 : r2 = 4 : 3
4πr12 : 4π22
42 : 32
16 : 9

Question 12.
A cylinder, a cone and a hemisphere have same base and same height, then the ratio of their volumes = ……………………
Solution:
Vcylinder : Vcone : Vhemisphere
πr2h : $$\frac{1}{3}$$πr2h : $$\frac{2}{3}$$πr3
πr3 : $$\frac{1}{3}$$πr3 : $$\frac{2}{3}$$πr3 (∵ r = h)
1 : $$\frac{1}{3}$$ : $$\frac{2}{3}$$
3 : 1 : 2

Question 13.
In a cone, l2 = ……………
h2 + r2

Question 14.
In a cone l2 – h2 = ……………..
r2

Question 15.
Foot ball is an example of …………………
Sphere

Question 16.
Vertical cross section of cylinder is …………..
A) Square
B) Circle
C) Semi circle
D) Rectangle
D) Rectangle

Question 17.
The edge of a cube is 12 cm then its volume is …………… cm3.
Solution:
V = 123 = 1728 cm3

Question 18.
The volume of cube is 125 cm3 then its edge is …………….. cm.
5

Question 19.
Joker’s cap is an example of ……………….
A) Cuboid
B) Semi circle
C) Sphere
D) Cone
Solution:
D) Cone

Question 20.
Heap of stones is an example of ……………..
A) Cone
B) Cylinder
C) Circle
D) None
A) Cone

Question 21.
Volume of cone = …………… $$\bar{x}$$ Volume of cylinder.
$$\frac{1}{3}$$

Question 22.
Stem of a tree is an example of …………………
A) Sphere
B) Cone
C) Cylinder
D) Rectangle
C) Cylinder

Question 23.
Diagonal of a cube of edge ‘a’ units is …………….
Solution:
a√3

Question 24.
The volume of a right circular cylinder of base radius 7 cm and height 10 cm is …………… cm3.
Solution:
r = 7 cm
h = 10 cm
V = πr2h
= $$\frac{22}{7}$$ × 7 × 7 × 10
= 1540 cm3

Question 25.
The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3 then the ratio of their volumes =………………..
r1 : r2 = 2 : 3
h1 : h2 = 5 : 3
πr12h1 : πr22h2
22 × 5 : 32 × 3
20 : 27

Question 26.
The radius (r) of a sphere is reduced to its half then new volume would be ………………..
Solution:
Let radius = $$\frac{r}{2}$$
V = $$\frac{r}{2}$$π($$\frac{r}{2}$$)3 = $$\frac{4}{3}$$π
$$\frac{r^3}{8}$$
= $$\frac{1}{8}$$($$\frac{4}{3}$$πr3)

Question 27.
Find the curved surface area of a cone whose radius Is one-third of its height.
Solution:
l = $$\sqrt{h^2+r^2}$$
= $$\sqrt{\left(\frac{\mathrm{h}}{3}\right)^2+\mathrm{h}^2}$$ = $$\frac{\sqrt{10} h}{3}$$
CSA = πrl = $$\frac{\sqrt{10} \pi \mathrm{h}^2}{9}$$

Question 28.
The surface areas of two hemispheres are in the ratio 25 : 49. Then find the ratio of their radius.
Solution:
$$\frac{3 \pi r^2}{3 \pi R^2}$$ = $$\frac{25}{49}$$ ⇒ $$\frac{\mathrm{r}^2}{\mathrm{R}^2}$$ = $$\frac{25}{49}$$ ⇒ $$\frac{r}{R}$$ = $$\frac{5}{7}$$ = 5 : 7

Question 29.
The total surface area of sphere is 98.56 cm2 then the radius of the sphere is ………………… cm.
Solution:
4πr2 = 98.56
r2 = $$\frac{98.56 \times 7}{4 \times 22}$$
r2 = 7.84
r = 2.8 cm

Question 30.
What is the volume nearly equal to a metallic spherical ball of radius 4.5 cm?
Solution:
V = $$\frac{4}{3}$$ πr3 ⇒ V = $$\frac{4}{3}$$ × $$\frac{22}{7}$$ × 4.5 × 4.5 × 4.5
V = 382 cm3

Question 31.
If the volume and surface area of a sphere is numerically equal, then its radius is ……………. units.
Solution:
4πr3 = $$\frac{4}{3}$$πr3 ⇒ 1 = $$\frac{1}{3}$$ × r
r = 3 units

Question 32.
The surface area of sphere of radius 10.5 cm is …………….. cm2.
Solution:
Surface area = 4πr2
= 4 × $$\frac{22}{7}$$ × 10.5 × 10.5 cm2
= 1386 cm2

Question 33.
The surface area of asphere of diameter 3.5 cm is ………………. cm2.
Solution:
r = $$\frac{3.5}{2}$$ cm
Surface area of sphere = 4πr2
4 × $$\frac{22}{7}$$ × $$\frac{3.5}{2}$$ × $$\frac{3.5}{2}$$
= 38.5 cm2

Question 34.
The surface area of a sphere is 154 cm2 then its radius is ……………. cm.
Solution:
4πr2 = 154
4 × $$\frac{22}{7}$$ × r2 = 154
r2 = $$\frac{154 \times 7}{22 \times 4}$$
r2 = 12.25
r = 3.5 cm

Question 35.
In a cone, the diameter ofbase is 10.5 cm and slant height is 10 cm then its curved surface area is……….. cm2.
Solution:
r = $$\frac{10.5}{2}$$
l = 10 cm
CSA = πrl
= $$\frac{22}{7}$$ × $$\frac{10.5}{2}$$ × 10
= 165 cm2

Question 36.
When two cubes are joined together the resultant figure is
A) Cone
B) Cuboid
C) Cube
D) Cylinder
B) Cuboid

Question 37.
TSA of cone = …………………..
πr (l + t)

Question 38.
Curved surface area of a hemisphere is 308 cm2 then its total surface area is…………. cm2.
Solution:
2πr2 = 308
2 × $$\frac{22}{7}$$ × r2 = 308
r2 = 308 × $$\frac{7}{22}$$ × $$\frac{1}{2}$$
r2 = 49
r = 7 cm
TSA = 3πr2
= 3 × $$\frac{22}{7}$$ × 7 × 7
= 66 × 7
= 462 cm2

Question 39.
Write the area of four walls of a room.
2h (l + b)

Question 40.
Two cubes of edge 6 cm are joined end to end. Then the resulting cuboid has the total surface area ……………… cm2.
A) 160
B) 360
C) 240
D) None
Solution:
B) 360
l = 6 + 6 = 12 cm
b = 6 cm
h = 6 cm
TSA = 2(lb + bh + lh)
= 2(180)
= 360 cm2

Question 41.
Assertion (A) : The volume of a cube of edge 1 cm is 2 cm3.
Reason (R) : Volume of a cube is a3
A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R) istrue.
D) Assertion (A) is false but reason (R) istrue.

Question 42.
Assertion (A) : Area of four walls of a room is 2h(l + b).
Reason (R) : CSA of cone = πrl.
A) Both assertion (A) and Treason (R) are true and reason (R) is the correct explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R) is true.
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

Question 43.
Assertion (A): Volume of cylinder = πr2h.
Reason (R) : A cube has 6 faces. .
A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R) is true.
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

Question 44.
Assertion (A) : Volume of hemisphere of radius 7 cm is $$\frac{616}{3}$$ cm2.
Reason (R): Volume of hemisphere = $$\frac{2}{3}$$πr3
A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R)

Question 45.
Assertion (A) : CSA of cylinder with h = 7 cm, r = 10 cm is 440 cm2
Reason (R) : Volume of cube = a3
A) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R) is true.
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

Question 46.
Assertion (A): The maximum length of a rod that can be placed in a room of dimensions 12 m × 11 m × 10 m is 13 m.
Reason (R) : Diagonal of a cuboid is $$\sqrt{l^2+b^2+h^2}$$
A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R) is true.
D) Assertion (A) is false but reason (R) is true.

Question 47.
Assertion (A) : The volume of sphere of radius 5 cm is 4π cm3
Reason (R) : Volume of sphere = πr3.
A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R) is true.
D) Assertion (A) is false but reason (R) is true.

Question 48.
Assertion (A) : In a cone h = 4 cm r = 3 cm then l = 5 cm
Reason (R) : In a cone l2 = h2 + r2
A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R) is true.
A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question 49.
Assertion (A) : In a hemisphere if r = 10.5 cm then volume = 95.2 cm3
Reason (R): Volume of Hemisphere is $$\frac{2}{3}$$ πr3
A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R) is true.
D) Assertion (A) is false but reason (R) is true.

Question 50.
Assertion (A) : The ratio of volume of two cubes is 1 : 27 then the ratio of their surface areas is 1 : 9
Reason (R) : Surface area of cube = 2h (l + b) – 2
A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R:) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R) is true.
B) Both assertion (A) and reason (R) are true but reason (R:) is not the correct explanation of assertion (A).

Question 51.
The capacity of an oil drum is 10 litres then what is its volume ? ?(in cm3)
10,000 cm3

Question 52.
Food grains are to be stored in containers of the same base length and height. Which type of containers are required less in number to store a fixed quantity of grains ?
i) Right Circular Cylinder
ii) Cube
iii) Right Circular Cone
ii) Cube

Question 53.
Choose the correct answer satisfying the following statements.
Statement (A) : The ratio of volumes of cone and cylinder of same base and same height is 3 : 1 Statement (B) : Hie ratio of volumes of sphere and cone of same radius and same height is 2 : 1
i) Both A and B are true
ii) A is true, B is false
iii) A is false, B is true
iv) Both A and B are false
iv) Both A and Bare false

Question 54.
Find the volume of a cylinder whose base radius is 3 cm and height is 7 cm.
Solution:
Radius of a cylinder r = 3 cm.
Height h = 7 cm.
Volume of the cylinder = πr2h
= $$\frac{22}{7}$$ × (3)2 × 7
= $$\frac{22}{7}$$ × 9 × 7
= 198 cm3

Question 55.
Curved surface area of Cone is πrl where l stands __________ .
l = Slant height

Question 56.
The volume of a cube whose edge is 6 cm is
A) 108 cm3
B) 18 cm3
C) 216 cm3
D) 144 cm3
C) 216 cm3

Question 57.
Find the volume of a cube, whose side is 4 cm.
Solution:
Side = a = 4 cm
Volume of cube V = a3 = 43 = 64 cm3

Question 58.
Find the volume of a sphere of radius 21 cm. (Take π = 22/7)
Solution:
Volume of the sphere = $$\frac{4}{3}$$ πr3
= $$\frac{4}{3}$$ × $$\frac{22}{7}$$ × 21 × 21 × 21
= 38, 808 (cm)3

Question 59.
Find the total surface area of a hemisphere, whose radius is 7 cm.
Solution:
Radius of Hemisphere, r = 7 cm
T.S.A of Hemisphere = 3πr2
= 3 × $$\frac{22}{7}$$ × 7 × 7
= 462 (cm)2

Question 60.
Find the volume of right circular cone with radius 3 cm. and height 14 cm.
Solution:
Volume of right circular cone = $$\frac{1}{3}$$πr2 h
= $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 3 × 3 × 14
= 132 cm2

### 10th Class Maths Surface Areas and Volumes 2 Mark Important Questions

Question 1.
From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of same height and same base radius is removed. Find the volume of the remaining solid.
Solution:
In a cylinder,
height, h = 14 cm
In a case, h = 14 cm
r = 6 cm
Volume of remaining solid
= Vcylinder – Vcone
= πr2h – $$\frac{1}{3}$$πr2h
= $$\frac{3 \pi r^2-\pi r^2 h}{3}$$ = $$\frac{2 \pi r^2 h}{3}$$
= $$\frac{2 \times 22 \times 6 \times 6 \times 14}{7 \times \times 3}$$
= 44 × 24
= 1056 cm3

Question 2.
Find the curved surface area of a right circular cone whose height is 15 cm and base radius is 8 cm.
[Use π = $$\frac{22}{7}$$]
Solution:
In a cone, height, h = 15 cm
l = h2 + r2
= 152 + 82
= 225 + 64
l2 = 289
l = $$\sqrt{289}$$
l = 17 cm
CSA = πrl = $$\frac{22}{7}$$ × 8 × 7 cm2
= $$\frac{2992}{7}$$ cm2

Question 3.
The surface area of a sphere is 616 sq cm. Find its radius. [Use π = $$\frac{22}{7}$$]
Solution:
Surface area of sphere = 616 cm2
4πr2 = 616
4 × $$\frac{22}{7}$$ × r2 = 616
r2 = 616 × $$\frac{7}{20}$$ × $$\frac{1}{4}$$
r2 = 49
r = 7 cm
∴ Radius of sphere = 7 cm

Question 4.
Three cubes each of volume 64 cm3 are joined end to end to form a cuboid. Find the total surface area of the cuboid so formed ?
Solution:
Volume of cube = 64 cm3
a3 = 64
a3 = 43
a = 4 cm
When 3 cubes are joined together cuboid is formed.
l = 4 + 4 + 4 = 12 cm
b = 4 cm
h = 4 cm
Total surface area = 2(lb + bh + lh)
= 2(12 × 4 + 4 × 4 + 12 × 4)
= 2(481 + 6 + 48)
= 2(112) cm2
= 224 cm2

Question 5.
The volume of a right circular cylinder with its height equal to the radius is 25$$\frac{1}{7}$$
cm3. Find the height of the cylinder. (Use π = $$\frac{22}{7}$$)
Solution:
In a cylinder, r = h.
Volume of cylinder = πr2h
25$$\frac{1}{7}$$ = πr3 ⇒ $$\frac{176}{7}$$ = $$\frac{22}{7}$$r3
8 = r3
r3 = 23
r = 2 cm
∴ height h = 2 cm

Question 6.
A cubical block of side 7 cm is surmounted by a hemisphere of largest possible diameter as shown in Figure. Find the total surface area of the solid.
Solution:

Side of cube = 7 cm
radius of hemisphere = $$\frac{7}{2}$$ cm
Total surface area of solid
= TSA of cube + Surface area of hemisphere – Area of circle
= 6a2 + 2πr2 – πr2
= 6(7)2 + πr2
= 294 + $$\frac{22}{7}$$ × $$\frac{7}{2}$$ × $$\frac{7}{2}$$
= 294 + $$\frac{77}{2}$$ cm2
= 294 + 38.5 cm2
= 332.5 cm2

Question 7.
A sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with water. The dimater of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Solution:
Let the water level raised in cylindrical vessel be h cm
Vsphere = Vwater displaced in cylinder
$$\frac{4 \pi r^3}{3}$$ = πR2h
$$\frac{4}{3}$$π(3)3 = π(6)2h
$$\frac{4}{3}$$ × 27 = 36 h
h = $$\frac{36}{36}$$
h = 1 cm

Question 8.
A cylinder and a cone have base radii 5 cm and 3 cm respectively and their respective heights are 4 cm and 8 cm. Find the ratio of their volumes.
Solution:
Cone
r = 3 cm
h = 8 cm
Volume = $$\frac{1}{3}$$ πr2h
= $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 3 × 3 × 8

Cylinder
r = 5 cm
h = 4 cm
Volume = πr2h
= $$\frac{22}{7}$$ × 5 × 5 × 4
Ratio = $$\frac{22}{7}$$ × 3 × 8 : $$\frac{22}{7}$$ × 5 × 5 × 4
= 3 × 8 : 5 × 5 × 4 = 6 : 25

Question 9.
A sphere of maximum volume is cut from a solid hemisphere of radius 6 cm. Find the volume of the cut out sphere.
Solution:
In Hemisphere, r = 6 cm
radius of sphere, R = $$\frac{6}{2}$$ = 3 cm
Volume of sphere = $$\frac{4}{3}$$πR3
= $$\frac{4}{3}$$ × $$\frac{22}{7}$$ × 3
= $$\frac{4}{3}$$ × 9 = $$\frac{729}{7}$$
= 113.14 cm3

Question 10.
Two cubes of 5 cm each are kept together joining edge to edge to form a cuboid. Find the surface area of cuboid so formed.
Solution:
Two cubes of each edge 5 cm are joined then it is a cuboid
l = 5 + 5 = 10 cm
b = 5 cm
c = 5 cm
Surface of cuboid = 2(lb + bh + lh)
= 2[ 10 × 5 + 5 × 5 + 10 × 5]
= 2[50 + 25 + 50]
= 2(125) cm2
= 250 cm2

Question 11.
If the total surface area of a solid hemisphere is 462 cm2, find its volume [Take π = $$\frac{22}{7}$$]
Solution:
Total surface area of hemisphere = 462 cm2
3πr2 = 462
3 × $$\frac{22}{7}$$ × r2 = 462
r2 = 462 × $$\frac{7}{22}$$ × $$\frac{1}{3}$$
r2 = 21 × 7 × $$\frac{1}{3}$$
r2 = 7 × 7
r = 7 cm
Volume = $$\frac{2}{3}$$ πr3
= $$\frac{2}{3}$$ × $$\frac{2}{3}$$ × 7 × 7 × 7 = $$\frac{2156}{3}$$ cm3

Question 12.
A 5m wide cloth is used to make a conical tent of base diameter 14m and height 24m. Find the cost of cloth used at the rate of ₹ 25 per metre.
Solution:
In a cone, h = 24 m
d = 14 m
r = $$\frac{14}{2}$$ = 7 m
l2 = h2 + r2
l2 = 242 + 72
l2 = 576 + 49
l2 = 625
l = $$\sqrt{625}$$
l = 25 m
CSA of cone = πrl
= $$\frac{22}{7}$$ × 7 × 25 cm2
= 550 m2
Length of cloth used = $$\frac{\text { Area }}{\text { width of cloth }}$$ = $$\frac{550}{5}$$ – 110 m
Cost of cloth = ₹ 25 per m.
Total cost = ₹ 25 × 110
= ₹ 2750

Question 13.
A conical tent is 10m high and the radius of its base is 24m. Find the cost of cloth used to make the tent at ₹ 70/m3.
Solution:
In a cone, h = 10 m
r = 24 m
CSA = πrl
l = $$\sqrt{\mathrm{h}^2+\mathrm{r}^2}$$
= π × 24 × $$\sqrt{10^2+24^2}$$
= $$\frac{22}{7}$$ × 24 × $$\sqrt{676}$$
= $$\frac{22}{7}$$ × 24 × 26 m2
Cost = ₹ 70 × $$\frac{22}{7}$$ × 24 × 26
= ₹ 137280

Question 14.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the volume of the remaining solid to the nearest cm3 (Use π = $$\frac{22}{7}$$)
Solution:
Height of cone = height of cylinder, h = 2.4 cm
Diameter of cylindrical part = 1.4 cm
r = $$\frac{1.4}{2}$$ = 0.7cm
Slant heigth (l) = $$\sqrt{\mathrm{h}^2+\mathrm{r}^2}$$
= $$\sqrt{(2.4)^2+(0.7)^2}$$
= $$\sqrt{5.76+0.49}$$
= $$\sqrt{6.25}$$
= 2.5 cm
Total surface area of remaining solid
= CSA of cylindrical part + CSA of conical part + Area of cylindrical base
= 2πrh + πrl + πr2
= 2 × $$\frac{22}{7}$$ × 0.7 × 2.4 × $$\frac{22}{7}$$ × 0.7 × 2.5 + $$\frac{22}{7}$$ × 0.7 × 0.7 cm2
= 4.4 × 2.4 + 2.2 × 2 + 2.2 × 0.7
= 10.56 + 5.50 + 1.54
= 17.60 cm2
∴ TSA of remaining solid in nearest to 18 cm2.

Question 15.
The circumference of the base of a 9 in high wooden solid cone is 44 m. Find the volume of the cone.
Solution:
h = 9 m
Circumference of base = 2πr = 44 m
2 × $$\frac{22}{7}$$ × r = 44
r = 44 × $$\frac{7}{22}$$ × $$\frac{1}{2}$$
r = 7 m
Volume of cone = $$\frac{1}{3}$$ πr2h
= $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 7 × 7 × 9
= 22 × 21 m2
= 462 m3

Question 16.
Find the number of coins of 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:
Let n be the number of cones required Volume of cylinder = n × Volume of 1.5 cm diameter coin
πR2H = n × πr2h ⇒ n = $$\frac{R^2 H}{r^2 h}$$
= $$\frac{\left(\frac{4.5}{2}\right)^2 \times 10}{\left(\frac{1.5}{2}\right)^2 \times \frac{2}{10}}$$
= $$\frac{45}{20}$$ × $$\frac{45}{20}$$ × $$\frac{10 \times 20 \times 20 \times 10}{15 \times 15 \times 2}$$ = 9 × 10 × 5
n = 450.
∴ number of coins required = 450.

Question 17.
A glass cylinder with diameter 20 cm has water to a height of 9 cm. A metal cube of 8 cm edge is immersed in it completely. Calculate the height by which water will rise in the cylinder. [Use π = 3.14]
Solution:
Let the water rise be h cm
Volume of water displaced = Volume of the cube of edge 8 cm
πr2h = 83
3.14 × 102 × h = 8 × 8 × 8
h = $$\frac{8 \times 8 \times 8}{3.14 \times 10 \times 10}$$
h = $$\frac{512}{314}$$ ⇒ h = 1.6 cm
∴ height = 1.6 cm

Question 18.
In a construction of kitchen shed at school, a truck unloaded the sand which was formed in the shape of a cone. The base radius of the cone is 2.7 m. and its height is 7 m. Find the volume of sand unloaded there.
Solution:
Base radius of the cone (r) = 2.7 m
height (h) = 7 m.
Volume of the cone = $$\frac{1}{3}$$pr2h
= $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × (2.7)2 × 7
= $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 2.7 × 2.7 × 7
∴ Volume of the sand = 53.46 m3.

### 10th Class Maths Surface Areas and Volumes 4 Mark Important Questions

Question 1.
Find the number of spherical lead shots, each of diameter 6 cm that can be made from a splid cuboid of lead having dimensions 24 cm × 22 cm × 12 cm.
Solution:
In a cuboid l = 24 cm
b = 22 cm
h = 12 cm
Spherical shot, d = 6 cm
r = $$\frac{6}{2}$$ = 3 cm

Question 2.
A wooden souvenir is made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm then find the total cost of polishing the souvenir at the rate of Rs. 10 per cm2.
Solution:
In cylinder, h = 10 cm
r = 3.5 cm
TSA of the article = CSA of cylinder + 2
surface area of hemisphere
= 2πrh + 2(2πr2)
= 2πr (h + 2r)
= 2 × $$\frac{22}{7}$$ × 3.5(10 + 2(3.5))cm2
= 2 × 22 × 0.5 (10 + 7) cm2
= 22(17)
= 374 cm2
Cost of polishing = ₹ 10 per cm2
Total cost = 374 × ₹ 10
= ₹ 3740

Question 3.
A right circular cone of radius 3 cm has a curved surface area of 47.1 cm2. Find the volume of the cone (Use π = 3.14)
Solution:
In a cone raius = 3 cm
surface area = 47.1 cm2
πrl = 47.1
3.14 × 3 × l = 47.1
l = $$\frac{47.1}{3.14 \times 3}$$
l = $$\frac{15}{3}$$
l = 5 cm
Volume of cone = $$\frac{1}{3}$$πr2h
l2 = h2 + r2
52 = h2 + 32
25 – 9 = h2
h2 = 16
h = 4 cm
Volume = $$\frac{1}{3}$$ × 3.14 × 3 × 3 × 4 cm3
= 3.14 × 12 cm3
= 37.68 cm3

Question 4.
The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 sq.cm. Find the volume of the cylinder (Use π = $$\frac{22}{7}$$
Solution:
In a cone,
r + h = 37
TSA = 16.28 cm2
2πr(h + r) = 1628
2 × $$\frac{22}{7}$$ × r × 37 = 1628
r = 1628 × $$\frac{7}{22}$$ × $$\frac{1}{37}$$ × $$\frac{1}{2}$$
22 37 2
r = 7 cm
h = 37 – 7 = 30 cm
Circumference = 2πr = 2 × $$\frac{22}{7}$$ × 7 = 44 cm
Volume of cylinder = πr2h
= $$\frac{22}{7}$$ × 7 × 7 × 30m3
= 22 × 210 cm3
= 4620 cm3

Question 5.
In the given figure, a tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3m respectively and the slant height of conical part is 2.8m. Find the cost of canvas needed to make the tent if the canvas available at the rate of ₹ 500 per sq. (Use π = $$\frac{22}{7}$$)

Solution:
r = $$\frac{3}{2}$$ m
Slant height, l = 2.8 m
Height of cylinder, h = 2.1 m ,
Canvas needed to make the tent = CSA of conical part + CSA of cylindrical part
= πrl + 2πrh = $$\frac{22}{7}$$ × $$\frac{3}{2}$$ × 2.8 + 2 × $$\frac{22}{7}$$ × $$\frac{3}{2}$$ × 2.1
= 13.2 + 19.8 = 33 m2
Cost of 1 m2 canvas = ₹ 500
Cost of 33 m2 convas = 33 × 500
= ₹ 16,500
∴ Cost to make the tent = 116,500

Question 6.
A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the
hemisphere can have ? Find the cost of painting the total surface area of the solid so formed, at the rate of ₹ 5 per 100 sq. cm. (Use ₹ = 3.14)
Solution:

Total surface area of the solid = TSA of cube + CSA of hemisphere – Area of base of hemisphere
= 6a2 + 2πr2 – πr2
= 6 × 102 + 2 × 3.14 × 52 – 314 × 52cm2
= 600 + 157 – 78.5
= 678.5 cm2
Cost of painting = 5 per 100 cm2
Cost of painting the solid
= 678.5 × $$\frac{5}{100}$$
= ₹ 33.925

Question 7.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of some radius on its circular face. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Total surface area of toy = CSA of cone + Surface area of hemisphere
= πrl + 2πr2
In a cone, r = 3.5 cm,
h = 15.5 – 3.5 = 12 cm
l2 – h2 + r2
l2 = 122 + 3.52
l2 = 144 + 12.25
l2 = 156.25
l = $$\sqrt{156.25}$$
l = 12.5

Question 8.
A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of dimeter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer.
Solution:
In hemisphere
Diameter, d = 36 cm
Volume of spherical bowl = $$\frac{2}{3}$$πr3
= $$\frac{2}{3}$$ × $$\frac{22}{7}$$ × 18 × 18 × 18
= 12219.43 cm3
In cylinder,
r = 3 cm
Volume = πr2h
Volume of 72 bottles = $$\frac{22}{7}$$ × 3 × 3 × h × 72
Now 10% of 12219.43 = 1221.943 cm3 is wasted.
Remaining volume
= 12219.43 – 1221.943 cm3
= 10997.487 cm3
72 × $$\frac{22}{7}$$ × 3 × 3 × h = 10997.487
h = $$\frac{10997.487 \times 7}{72 \times 22 \times 9}$$
height, h = 5.4 cm

Question 9.
A hemispherical bowl of internal diameter 30 cm contains some liquid. This liquid is to be filled into cylindrical shaped bottles each of diameter 5 cm and height 6 cm. Find the number of bottles necessary to empty the bowl.
Solution:
Inner diameter of bowl = 30 cm
Inner radius of bowl = 15 cm
Volume of liquid = $$\frac{2}{3}$$πr3
= $$\frac{2}{3}$$ × π × 15 × 15 × 15 cm3
Radius of each cylindrical bottle = 2.5 cm
h = 6 cm
Volume of each cylindrical bottle
= πr2h
= π × $$\frac{5}{2}$$ × $$\frac{5}{2}$$ × 6 cm3
= $$\frac{25}{4}$$ × 6π
Required number of bottles
= $$\frac{\frac{2}{3} \times \pi \times 15 \times 15 \times 15}{\frac{25}{2} \times \pi}$$ = 60

Question 10.
The rain water from 22 m × 20 m roof drains into cylindrical vessel of diameter 2 m and height 3.5 m. If the rain water collected from the roof fills $$\frac{4}{5}$$th of cylindrical vessel then find the rainfall in cm.
Solution:
Length of the roof = 22 m
Breadth of the roof = 20 m
Let the height of water on the roof be h m
Radius of cylinderical vessel = r = $$\frac{2}{2}$$ = 1m
Height of water in the cylindrical vessel H = 3.5 m
Volume of water falling on the roof = Volume of water in the cylinder
lbh = πr2h
22 × 20 × h = π × 72 × 3.5
h = $$\frac{22}{7}$$ × $$\frac{1 \times 35}{22 \times 20}$$
h = $$\frac{1}{40}$$ m
h = 2.5 cm
∴ 2.5 cm of rainfall falls on the roof.

Question 11.
A hollow cylindrical pipe is made up of copper. It is 21 m long, the outer and inner diameters of the pipe are 10 cm and 6 cm respectively. Find the volume of the copper used in making the pipe.
Solution:
Height of cylindrical pipe, h = 21 m = 210 cm
External Radius, R = $$\frac{10}{2}$$ = 5 cm
Internal Radius, r = $$\frac{6}{2}$$ = 3cm
Volume of copper making the pipe
= πR2h – πr2h
= πh(R2 – r2)
= $$\frac{22}{7}$$ × 210(52 – 32)
= $$\frac{22}{7}$$ × 210(25 – 9)
= 22 × 30 × 16
= 10560 cm3

Question 12.
A glass is in the shape of a cylinder of radius 7 cm and height 10 cm. Find the volume of juice in litre required to fill 6 such glasses. (Use π = $$\frac{22}{7}$$)
Solution:
height =10 cm
volume of 1 gram = πr2h
= $$\frac{22}{7}$$ × 7 × 7 × 10cm3
= 22 × 70 cm3
= 1,540 cm3
∴ Volume of 6 such glasses = 1540 × 6 = 9,240 cm3

Question 13.
The radii of two cylinder are in the ratio of 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their curved surface area.
Solution:
In two cylinders
Ratio in radii of two cylinder = 2 : 3
Ratio of their heights = 5 : 3
r1 : r2 = 2 : 3
h1 : h2 = 5 : 3
$$\frac{r_1}{r_2}$$ = $$\frac{2}{3}$$, $$\frac{h_1}{h_2}$$ = $$\frac{5}{3}$$
Ratio of curved surface area
$$\frac{2 \pi r_1 h_1}{2 \pi r_2 h_2}$$ = $$\frac{r_1 h_1}{r_2 h_2}$$
$$\frac{r_1 h_1^j}{r_2 h_2}$$ = $$\frac{10}{9}$$
∴ Required Ratio = 10 : 9.

Question 14.
A wooden toy was made by scooping at a hemisphere of same radius from each end of a solid cylinder. If the length of the cylinder is 10 cm, then find the volume of that toy.
Solution:

Radius of cylinder = Radius of hemisphere = r = 3.5 cm
height of the cylinder = h = 10 cm
Volume of the toy = Volume of the cylinder – Volume of two hemisphere
= πr2h – 2 × $$\frac{2}{3}$$πr3 = πr2 (h – $$\frac{4 r}{3}$$)
= πr2 (10 – $$\frac{4 \times 3.5}{3}$$) = πr2 (10 – $$\frac{14}{3}$$)
= $$\frac{22}{7}$$ × 3.5 × 3.5 × $$\frac{16}{3}$$
= $$\frac{616}{3}$$ cm3
= 205.333 cm3

Question 15.
The base area of a cone is 616 sq.cm and its height is 48 cm. Find its total surface area.
Solution:

The base area of a cone = πr2 = 616 cm2
Given height of cone = 48 cm = h
Now π2 = 616
⇒ $$\frac{22}{7}$$ × r2 = 616
⇒ r2 = $$\frac{616 \times 7}{22}$$
= 7 × 4 × 7
⇒ r = $$\sqrt{7 \times 4 \times 7}$$
= 7 × 2
= 14 cm
If r = 14, h = 48, then
Slant height (l) = $$\sqrt{r^2+h^2}$$
= $$\sqrt{14^2+48^2}$$
= $$\sqrt{196+2304}$$
= $$\sqrt{2500}$$
= 50 cm.
∴ Curved surface area of the cone
= πrl = $$\frac{22}{7}$$ × 14 × 50 = 2200
∴ Total surface area
= C.S.A + base area
= 2200 + 616 = 2816 cm2

### 10th Class Maths Surface Areas and Volumes 8 Mark Important Questions

Question 1.
The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform cross-section. If the length of the wire is 36 m, And its radius.
Solution:
Given diameter of sphere = 6 cm
Radius of the sphere r = $$\frac{6}{2}$$ = 3 cm
length of wire h = 36 m = 3600 cm
Volume of sphere V = $$\frac{4}{3}$$πr3
= $$\frac{4}{3}$$π × 3 × 3 ⇒ V = 36π cm3
but, given sphere is melted and drawn into a uniform wire. Wire Is in the shape of cylinder.
So, volume of wire = volume of sphere.
πr2h = 36π
π × r2 × 3600 = 36π
r2 = $$\frac{36 \pi}{\pi \times 3600}$$ = $$\frac{1}{100}$$ = ($$\frac{1}{10}$$)2
r = $$\frac{1}{10}$$ = 0.1 cm = 1 mm
Therefore radius of wire = 0.1 cm/1 mm

Question 2.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Given radius of the sphere r = 4.2 cm
radius of the cylinder = 6 cm
height of the cylinder = ?
metallic sphere is recast into cylinder.
So, volume of cylinder = volume of sphere
πr2h = $$\frac{4}{3}$$πr3
π . 6 × 6 × h = $$\frac{4}{3}$$π × 4.2 × 4.2 × 4.2
h = $$\frac{4}{3}$$ × $$\frac{\pi \times4.2 \times 4.2 \times4.2}{\pi \times 6 \times 6}$$
= 4 × 1.4 × 0.7 × 0.7 = 2.774 cm
Therefore, height of the cylinder = 2.774 cm.

Question 3.
A sphere of diameter 6 cm is dropped in a right circular cylinderical vessel pastly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Solution:
Given diameter of sphere = 6 cm
radius = $$\frac{d}{2}$$ = $$\frac{6}{2}$$ = 3 cm
diameter of the cylinder = 12 cm
radius = $$\frac{\mathrm{d}}{2}$$ = 6 cm
Volume of sphere = $$\frac{4}{3}$$πr3
= $$\frac{4}{3}$$π × 3 × 3 × 3
= 36π cm3
Volume of cylinder = πr2h
= π × 62 h cm3
But, Volume of cylinder = Volume of sphere
π × 36 × h = 36π
h = $$\frac{36 \pi}{\pi 36}$$ = 1cm
Therefore, water level raised by 1 cm.

Question 4.
Solid cylinder of brass 8 m high and 4 m diameter is melted and recast into a cone of diameter 3 m, find the height of the cone.
Given radius of cylinder (r) = $$\frac{\mathrm{d}}{2}$$ = $$\frac{4}{2}$$ = 2 m
height of cylinder h = 8 m
radius of cone r = $$\frac{3}{2}$$ m
Given volume of cone = volume of cylinder
$$\frac{1}{3}$$πr2h = πr2h
$$\frac{1}{3}$$ × π × $$\frac{3}{2}$$ $$\frac{3}{2}$$ × h = π × 2 × 2 × 8
h = $$\frac{\pi \times 2 \times 2 \times 8 \times 3 \times 2 \times 2 \times 2}{\pi \times 3 \times \times 3}$$
Therefore, height of the cone
$$\frac{128}{3}$$ = 42.66 m

Question 5.
Find the number of coins, 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:
Given coin is in the shape of cylinder
radius of coin = $$\frac{\mathrm{d}}{2}$$ = $$\frac{1.5}{2}$$ cm
Thickness of coin h = 0.2 cm
Volume of coin = πr2h
= π × $$\frac{1.5}{2}$$ × $$\frac{1.5}{2}$$ × 0.2 cm3
Radius of cylinder r = $$\frac{\mathrm{d}}{2}$$ = $$\frac{4.5}{2}$$ cm
height of cylinder h = 10 cm
Volume of cylinder = πr2h
= π × $$\frac{4.5}{2}$$ × $$\frac{4.5}{2}$$ × 10 cm3
Let number of coins = n
Coins are melted to form a cylinder.
So, n × volume of coin = volume of cylinder

Therefore number of coins = 450

Question 6.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Given diameter of the well d = 7 m
radius r = $$\frac{7}{2}$$ m
height of the well h = 20 m
length of the platform l = 22 m
height h = ?
Volume of the earth in the platform = Volume of deep well
l.b.h = πr2h

Therefore height of the platform = 2.5 m

Question 7.
A cylindrical pipe has inner diameter of 7 cm and water flows through it at 192.5 l per minute. Find the rate of flow in kilometres per hour.
Solution:
Given inner diameter of cylinder d = 7 cm
radius r = $$\frac{7}{2}$$ cm
Volume of water per minute = 192.5 l.
Volume of water per hour
= (192.5 × 60)l
= (192.5 × 60 × 1000) cm3
Let height of the cylinder = h cm
Volume of the cylinder = Volume of water flows per hour
πr2h = 192.50 × 60 × 1000
$$\frac{22}{7}$$ × $$\frac{7}{2}$$ × $$\frac{7}{2}$$ × h = 192.50 × 60 × 1000

= 2.5 × 60 × 1000 × 2
h = 300000 cm = 3 km.
Therefore, the rate of flow of water is 3 km per hour.

Question 8.
A copper wire 3 mm in diameter is wound about a cylinder whose length is 1.2 m and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of the copper wire to be 8.88 gram per cm.
Solution:
Given diameter of the wire = d = 3 mm = $$\frac{3}{10}$$ cm
Length of the wire (or) height of cylinder h = 1.2 m = 120 cm
Number of rounds taken by the wire in the cylinder = $$\frac{\text { length of wire }}{\text { diameter of wire }}$$
= $$\frac{\frac{120}{3}}{\frac{3}{10}}$$ = $$\frac{1200}{3}$$ = 400
Radius of the cylinder = $$\frac{\mathrm{d}}{2}$$ = $$\frac{10}{2}$$ cm
Circumference of wire per one round (or) length of wire per one round
= 2πr = 2 × 3.14 × $$\frac{10}{2}$$
= 2 × 31.4 × 5 = 3.14 cm
Total length of wire used to make a cylinder =
length of wire per one round x number of rounds = 31.4 × 400 = 12560 cm
Given density of copper wire per cm = 8.88 grams
Mass of the wire
= length of wire × density of copper per cm
= 12560 × 8.88 = 111532.8 grams
Therefore total mass of the wire
= 111.533 kg.

Question 9.
How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a cuboid 11 cm × 10 cm × 7 cm ?
Solution:
Given radius of the coin (r) = $$\frac{\mathrm{d}}{2}$$ = $$\frac{1.75}{2}$$ cm
thickness or height of coin h = 2 mm = $$\frac{2}{10}$$ cm
Volume of coin = πr2h

Let number of coins = n
Given coins are melted to form a cuboid dimensions of cuboid are 11 × 10 × 7
So, volume of x coins = volume of cuboid n × volume of coin = l.b.h
n × $$\frac{11 \times 0.25 \times 1.75}{10}$$ = 11 × 10 × 7
n = $$\frac{\times 11 \times 10 \times 7 \times 10}{\times 11 \times 0.25 \times 1.75}$$
Therefore number of coins n = 1600

Question 10.
Water in a canal 1.5 m wide and 6 m deep in flow with speed of 10 km/ hour. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired ?
Solution:
Given width of canal = 1.5 m
length of canal = 6 m
Area of the canal = 1.5 × 6 m2
Required area per 10 km/hour
= 1.5 × 6 × 10000
Required area per 10km/30minutes.
= 1.5 × 6 × 10000 × $$\frac{1}{2}$$
Standing water = 8 cm (or) × $$\frac{8}{100}$$ m
So, required area to irrigate = $$\frac{1.5 \times 6 \times 10000 \times \frac{1}{2}}{\frac{8}{100}}$$
= $$\frac{1.5 \times 6 \times 10000 \times 100}{8 \times 2}$$ m2
= 562500 m2

Question 11.
Cylindrical bucket 32 cm high and with radius of base 18 cm is filled with sand. This bucket is emptied out on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm. Find the radius and slant height of the heap.
Solution:

Given radius of cylinder r = 18 cm
height h = 32 cm
Volume of bucket = πr2h
= π × 18 × 18 × 32 cm3
Let radius of cone r = ?
height of cone h = 24 cm
Volume of cone = $$\frac{1}{3}$$πr2h
= $$\frac{1}{3}$$ × π × r2 × 24
= 8πr2 cm3
We know that volume of cone = volume of cylinder
8πr2 = π × 18 × 18 × 32
r2 = $$\frac{\pi \times 18 \times 18 \times 32}{8}$$
r2 = 36 × 36 = 362
radius of the cone r = 36 cm
Slant height l = $$\sqrt{r^2+h^2}$$ = 362 + 242
= $$\sqrt{1296+576}$$
= $$\sqrt{1872}$$ = 12$$\sqrt{13}$$ cm

Question 12.
The surface area of solid metallic sphere is 616 cm2. It is melted and recast into a cone of height 28 cm. Find the diameter of the base of the cone so formed.
Solution:
Given surface area of sphere
4πr2 = 616 cm2
4 × $$\frac{1}{3}$$ × r2 = 616
r2 = $$\frac{616 \times 7}{4 \times 22}$$
r2 = 72
So, radius of sphere r = 7 cm
Sphere is melted into cone,
Given height of cone h = 28 cm
radius of cone r = ?
that is volume of cone = volue of sphere

Diameter of cone = 2 × r = 2 × 7 = 14cm

Question 13.
A tent is in the form of a cylinder of diameter 20 m and height 2.5 m surmounted by a cone of equal base and height 7.5 m. Find the capacity of the tent and the cost of the canvas at Rs.100 per square metre.
Solution:

Given radius of the cylinder = $$\frac{\mathrm{d}}{2}$$ = $$\frac{20}{2}$$ = 10 m
height of the cylinder = 2.5 m
Volume of cylinder = πr2h
= π × 10 × 10 × 2.5 m3
radius of cone r = 10 m
height of cone h = 7.5 m
Volume of cone = $$\frac{1}{3}$$ πr2h
= $$\frac{1}{3}$$ × π × 10 × 10 × 7.5 m3
Capacity of tent = Volume of cylinder + Volume of cone
= π × 100 × 2.5 + $$\frac{1}{3}$$ π × 100 × 7.5
= 100π (2.5 + 2.5)
Capacity of tent = 100 × 3.14 × 5 = 1570 m3
Surface area of cone = πrl
l = $$\sqrt{10^2+7.5^2}$$
= $$\sqrt{100+56.25}$$
= $$\sqrt{156.25}$$
l = 12.5 m
πrl = π × 10 × 12.5 = 125π m2
Surface area of cylinder = 2πrh
= 2 × π × 10 × 2.5
= 50π m2
Total surface area of tent
= Surface area of cone + Surface area of cylinder
= 125π + 50π
= 175π m2
= 175 × 3.14
= 549.5 m2
Cost of canvas cloth of 100 per m2 is
= 549.5 × 100
= ₹ 54,950

AP 10th Class Maths Chapter 12 Important Questions Surface Areas and Volumes

Question 1.
Find the volume of a sphere of radius 21cm.(Take π = 22/7)
Solution:
Volume of the sphere = $$\frac{4}{3}$$ πr3
= $$\frac{4}{3}$$ × $$\frac{22}{7}$$ × 21 × 21 × 21 = 38,808(cm)3

Question 2.
Find the total surface area of a hemisphere, whose radius is 7 cm.
Solution:
Radius of Hemisphere, r = 7 cm
T.S.A of Hemisphere = 3πr2
= 3 × $$\frac{22}{7}$$ × 7 × 7 = 462 (cm)2

Question 3.
Find the volume of right circular cone with radius 3 cm. and height 14 cm.
Solution:
Volume of right circular cone = $$\frac{1}{3}$$ πr2h
= $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 3 × 3 × 14 = 132 cm3

Question 4.
Find the curved surface area of cylinder, whose radius is 7 cm. and height is 10 cm.
Solution:
Radius of a cylinder = r = 7 cm
Height of a cylinder = h – 10 cm
CSA of a cylinder = A = 2πrh
= 2 × $$\frac{22}{7}$$ × 7 × 10
= 44 × 10 = 440 sq.cm

Question 5.
The surface area of a football is 616 cm2, then find the radius of that ball. ( π = 22/7).
Solution:
Surface area of a football (sphere) = 4πr2.
4πr2 = 616
πr2 = 154
r2 = 154 × $$\frac{7}{22}$$
r2 = 7 × 7 = 49
∴ r = 7 cm

Question 6.
The volume of a cylinder is 448π cm3 and height is 7 cm. What is its radius?
Solution:
Volume of the cylinder = πr2h = 448π
Here h = 7cm ,r = r
then πr2 × 7 = 448π
r2 = $$\frac{448}{7}$$ = 64
∴ Radious (r) = 8 cm

Question 7.
Find the surface area of a sphere of radius 14 cm. (Take π = $$\frac{22}{7}$$)
Solution:
Radius of given sphere = (r) = 14 cm
Formula for surface area of sphere = 4πr2
∴ Surface area of given sphere = 4 × $$\frac{22}{7}$$ × 14 × 14 = 88 × 28 = 2464 cm2

Question 8.
Draw a rough diagram of a solid, showing the combination of a cone and cylinder, whose base radii are same.

AB is radius of cylinder as well as cone.

Question 9.
Find the area of the paper sheet required to make a conical joker cap with radius 7 cm and slant height 14 cm.
Solution:
Slant height = 14 cm
Area of paper sheet = πrl
= $$\frac{22}{7}$$ × 7 × 14
= 22 × 14 = 308 cm2

Question 10.
If the diameter of a sphere is equal to the side of a cube, find the ratio of their volumes.
Solution:
Let diameter of the sphere = d = side of the cube
∴ Radius of the sphere (r) = $$\frac{\mathrm{d}}{2}$$
volume of the sphere = $$\frac{4}{3}$$πr3
= $$\frac{4}{3}$$π($$\frac{\mathrm{d}}{2}$$)3 = $$\frac{4}{3}π$$$$\frac{\mathrm{d}^{3}}{8}$$ = $$\frac{\pi}{6}$$ d3
And, volume of the cube = d3
∴ The ratio of volume of sphere to volume of cube = $$\frac{\pi}{6}$$ d3 : d3 = $$\frac{\pi}{6}$$ : 1

Question 11.
Find the volume and total surface area of a hemisphere whose radius is 35 cm?
Solution:
Radius of the hemisphere (r) = 35 cm.
Volume of the hemisphere = $$\frac{2}{3}$$ πr3
= $$\frac{2}{3} \times \frac{22}{7}$$ × 35 × 35 × 35

Total surface area = 3πr2
= 3 × $$\frac{22}{7}$$ × 35 × 35 = 11550 cm2

Question 12.
A solid iron rod has a cylindrical shape. Its height is 11 cm and base diameter is 7 cm. Then find the total volume of 50 rods.
Solution:
The height of the cylinderical
Rod (h) = 11 cm,
Raidus (r) = 7/2 cm
Volume = πr2h = $$\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$$ × 11
= 423.5 (cm)3
Volume of 50 rods = 50 × 423.5 = 21175 (cm)3

Question 13.
Two cubes each of volume 125 cm3 are joined end to end together. Find the total surface area of the resulting cuboid.
Solution:
Volume of each cube = 125 cm3
∴ a3 = 125 cm3 ⇒ a = 5 cm

By joining these two cubes we get a cuboid, whose
length (l) = 10 cm
height (h) = 5 cm
then the total surface area
= 2 (lb + bh + lh)
= 2(10(5) + 5(5) + 5(10))
= 2(50 + 25 + 50)
= 2(125) = 250 cm2

Question 14.
The base area of a cone is 616 sq.cm and its height is 48 cm. Find its total surface area.
Solution:
The base area of a cone = πr2 = 616 cm2
Given height of cone = 48 cm = h

∴ Total surface area
= C.S.A + base area
= 2200 + 616 = 2816 cm2

Question 15.
Find the volume of largest right circular cone that can be cut out from a cube, whose edge is 7 cm.
Solution:
Volume of the cone (V) = $$\frac{1}{3}$$ πr2h
[r = $$\frac{s}{2}=\frac{h}{2}$$]
= $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × $$\frac{7}{2}$$ × $$\frac{7}{2}$$ × 7
∴ V = 89.83 cm3

Question 16.
If half of the vertical angle of a cone of height 3 cm is 60°. Find its volume.
Solution:

Let ‘B’ be the centre of base of the cone and BC is radius.
AB is vertical height (i.e.,)
AB = 3 cm
Given that ∠BAC = 60°
ΔABC is a right angled triangle, then
tan 60° = $$\frac{\mathrm{BC}}{\mathrm{AB}}$$
⇒ √3 = $$\frac{\mathrm{BC}}{3}$$ ⇒ BC = 3√3 cm
Now we have h = 3 cm, r = 3√3 cm
Volume of the cone = $$\frac{1}{3}$$ πr2h
$$\frac{1}{3}$$ π(3√3)2 × 3 = 27π cm3
(or)
= $$\frac{594}{7}$$ = 84.86 cm3

Question 17.
Find the volume and total surface area of a hemisphere of diameter 7 cm. Take π = $$\frac{22}{7}$$
Solution:
Diameter of hemisphere = 7 cm
∴ Radius of hemisphere = (r) = $$\frac{d}{2}=\frac{7}{2}$$ cm
Then volume of hemisphere = V = $$\frac{2}{3}$$πr3
= $$\frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}$$ = $$\frac{22 \times 49}{12}$$ = 89.83
Volume = 89.83 cm3
Total surface area of hemisphere = 3πr2
= 3 × $$\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$$ = $$\frac{77 \times 3}{2}$$ = 115.5 cm2

Question 18.
The length, breadth and height of a cuboid are (log 125 + log 8),
(log 1000 – log 10) and log 10 respectively. Find the total surface area of the cuboid.
Solution:
Length of cuboid (l) = (log 125 + log 8)
= (log 1000 – log 10)
Height of cuboid (h) = log 10
∴ (l) = log (125 × 8)
= log 1000
= log 103 = 3 log 10 = 3

(b) = log 1000 – log 10
= log $$\frac{1000}{10}$$
= log 100 = log 102 = 2
(h) = log 10 = 1
Then total surface area of cuboid
= 2(lb + bh + lh)
= 2(3 × 2 + 2 × 1 + 1 × 3)
= 2 (6 + 2 + 3) = 2(11)
= 22 square units.

Question 19.
The radius of a sphere is 3.5 cm. Find it’s surface area.
Solution:
Given, r = 3.5 cm
Surface area of sphere = 4πr2
= 4 × $$\frac{22}{7}$$ × 3.5 × 3.5
= $$\frac{88 \times 12.25}{7}$$ = 154 cm2

Question 20.
Express the volume of a cone in terms of volume of right circular cylinder of the same base and height and explain how you arrived at it.
Solution:
We know that,
Volume of cone = $$\frac{1}{3}$$ πr2h
Volume of cylinder = πr2h
Hence,
volume of cone : volume of cylinder
= $$\frac{1}{3}$$ πr2h : πr2h
Hence, volume of cone = $$\frac{1}{3}$$ × volume of cylinder.

Question 21.
The radius of a conical tent is 5m and its height is 12m. Calculate the length of the canvas used in making the tent if width of canvas is 2cm.
Solution:
Radius of the conical sent (r) = 5 m.
Height of the tent (h) = 12 m.
∴ Slant height of the cone
(l) = $$\sqrt{r^{2}+h^{2}}$$
= $$\sqrt{5^{2}+12^{2}}$$
= $$\sqrt{25+144}$$ = $$\sqrt{169}$$ 13 m.
Now, surface area of the tent = πrl
= $$\frac{22}{7}$$ × 5 × 13 = $$\frac{1430}{7}$$ m2
Area of the canvas used = $$\frac{1430}{7}$$ m2
It is given that the width of the canvas = 2m.
Length of the canvas used = $$\frac{\text { Area }}{\text { Width }}$$
= $$\frac{1430}{7} \times \frac{1}{2}$$
= 102.14 m

Question 22.
How many spherical balls can be made out of a solid cube of lead, whose edge measures 66 cm. and each ball being 3 cm in radius ?
Solution:
Side of the cube (s) = 66 cm
Radius of the spherical ball = r = 3 cm
Let the number of spherical balls that can be made = n
n × volume of a spherical ball = Volume of the cube
n × $$\frac{4}{3}$$πr3 = s3
n × $$\frac{4}{3} \times \frac{22}{7}$$ × 3 × 3 × 3 = (66)3
n = 66 × 66 × 66 × $$\frac{3}{4} \times \frac{7}{22} \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}$$
= 2541
Number of spherical balls can be made = 2541

Question 23.
An oil drum is in the shape of cylinder, whose diameter is 2 m and height is 7 m. The painter charges ? ₹ 5 per m2 to paint the drum. Find the total charges to be paid to the painter for
10 drums.
Solution:
It is given that diameter of the (oil drum) cylinder = 2 m.
Radius of cylinder = $$\frac{\mathrm{d}}{2}=\frac{2}{2}$$ = 1 m.
Total surface area of a cylindrical drum = 2πr(r + h)
= 2 × $$\frac{22}{7}$$ × 1(1 + 7)
= 2 × $$\frac{22}{7}$$ × 8 = $$\frac{352}{7}$$m2 = 50.28 m2.
So, the total surface area of a drum = 50.28 m2
Painting charge per 1 m2 = ₹ 5.
Cost of painting of 10 drums
= 50.28 × 5 × 10
= ₹ 2514

Question 24.
A hemisphere is cut out from one face of a cubical wooden block of edge 21 cm, such that the diameter of the hemisphere is equal to the edge of the Cube. Determine the total surface area of the remaining block.
Solution:
Let diameter of the hemisphere be ‘l’ = 21 cm
Radius of the hemisphere = $$\frac{l}{2}=\frac{21}{2}$$ cm
Also, length of the edge of the cube = l = 21 cm
∴ Total surface area of the remaining solid .

Question 25.
Two metallic spheres of radii 6 cm and 8 cm are melted along with another sphere and made into a big sphere of radius 12 cm. Find the radius of the third sphere.
Solution:
= r1 = 6 cm
r = 8 cm
Volume of the metallic spheres

⇒ 728 + x3 – 123
⇒ x3 = 123 – 728 = 1728 – 728
= 1000 cm
⇒ x = 10 cm.

Question 26.
Two cubes each of volume 216 cm3 are joined end to end together. Find the total surface area of the resulting cuboid.
Solution:
Given, volume of the cube V = a3 = 216 cm3
∴ a3 = 6 × 6 × 6 = 63
Hence a = 6 cm

When two cubes are added, the length of cuboid = 2a = 2 × 6 = 12 cm,
breadth = a = 6 cm,
height = a = 6 cm is formed.
∴ T.S.A. of the cuboid = 2(lb + bh +lh)
= 2(12×6 + 6×6 + 12×6)
= 2(72 + 36 + 72)
= 2 × 180 = 360 cm2
∴ The surface area of resulting cuboid is 360 cm2.

Question 27.
What is area of required cloth to make 10 conical hats having 7 cm ground radius and 24 cm height ?
Solution:

Ground radius of cone (r) = 7 cm
Height of cone (h) = 24 cm
Then its slant height
l = $$\sqrt{r^{2}+h^{2}}$$
= 72 + 242
= $$\sqrt{49+576}$$
l = $$\sqrt{625}$$ = 25 cm
Then C.S.A. of the cone = Area of cloth required
= πrl = $$\frac{22}{7}$$ × 7 × 25 = 550 cm2
Then the cloth required for 10 hats = 550 × 10 = 5500 cm2

Question 28.
A circle having 21 cm radius is cut into 3 equal parts to make 3 equal circular cones. Then what will be the radius of such cone ?
Solution:
Radius of the circle = 21 cm
Then the circumference of circle
= 2πr = 2 × $$\frac{22}{7}$$ × 21 = 132 cm
$$\frac{1}{3}$$ of this circumference of circle = Circumference of the ground of the
cone
= $$\frac{1}{3}$$ × 132 = 44 cm
Now let the ground radius of new cone = r
Then its ground circumference = 2πr = 44

So ground radius of cone = 7 cm.

Question 29.
Define “Regular cone”. Deduce formula for slant height of a regular cone.
Solution:

A solid having a circular base and vertex above the centre of circular base is called a regular cone.
A → Vertex
‘O’ centre of ground
OB is the radius of cone (r) and
0A is the height of cone (h)
AB is slant height (l)
Now ΔAOB is a right angled triangle.
Where ∠AOB = 90 and AB is hypotenuse.
Then from Pythagorus theorem
AB2 = OA2 + OB2 = r2 + h2
∴ AB = $$\sqrt{r^{2}+h^{2}}$$
∴ Slant hight (l) = $$\sqrt{r^{2}+h^{2}}$$

Question 30.
Draw a cone and label them.
Solution:

In the above cone
OB is the radius of the ground.
AO is the height of cone.
AB is slant height of cone.

Question 31.
Which kind of cones are formed by rotating on their axis of following triangles?
a) Equilateral
b) Right angled
c) Scalene
Solution:
a) Equilateral triangle : Having the altitude of this equilateral as rotat¬ing axis, on rotation we get a regular cone, (right cone)

b) Right angled triangle : Having a leg (a side other than hypotenuse) as rotating axis, on rotation of this right angled triangle we get a right circular cone.

c) Scalene triangle : Having a particular altitude from a vertex as rotating axis, on rotation we get a
cone

Question 32.
A sphere, a cylinder and a cone have the same radius and same height. Find the ratio of their volumes. (AS4) [Hint: Diameter of the sphere is equal to the heights of the cylinder and the cone.] A sphere, a cylinder and a cone have the same radius and same height.
Solution:
Sphere Cylinder Cone
Height = r 2r = h 2r = h

Now C.S.A of spehre = 4πr2 …………….(1)
C.S.A of cylinder = 2πrh
= 2πr.2r-2r = 4πr2 ……………….. (2)
Now for cone, slant height = l
= $$\sqrt{r^{2}+h^{2}}$$
= $$\sqrt{r^{2}+4 r^{2}}$$ ( ∵ h = 2r)
= $$\sqrt{5 r^{2}}$$ = √5(r)
Then C.S.A of cone = πrl = πr. √5r
= √5 πr2 ……………. (3)
∴ Ratio of C.S.A of sphere, cylinder and cone = 4πr2 : 4πr2 : √5 πr2
= 4 : 4 : √5

## AP 10th Class Maths Chapter 11 Important Questions Areas Related to Circles

These AP 10th Class Maths Chapter Wise Important Questions 11th Lesson Areas Related to Circles will help students prepare well for the exams.

## 11th Lesson Areas Related to Circles Class 10 Important Questions with Solutions

### 10th Class Maths Areas Related to Circles 1 Mark Important Questions

Question 1.
Find the circumference and area of a circle of radius 8.4 cm.
Solution:
Given radius of the circle = 8.4 cm
Circumference C = 2πr
= 2 × $$\frac{22}{7}$$ × 8.4
= 52.8 cm
Area of the circle A = πr2
= $$\frac{22}{7}$$ × 8.4 × 8.4
= 221.76 cm2

Question 2.
Find area of a circle whose circumference is 44 cm.
Solution:
Give circumference of the circle C = 44 cm
Circumference = 2πr = 44
= 2 × $$\frac{22}{7}$$ r = 44
r = $$\frac{44 \times 7}{2 \times 22}$$
r = 7 cm
Area of circle = πr2
= $$\frac{22}{7}$$ × 7 × 7 = 154 cm2

Question 3.
Define minor sector.
Solution:
A sector of a circle is called a minor sector if the minor arc of the circle is a part of the boundary.
Shaded part of AOB is the minor sector.

Question 4.
Define major sector.
Solution:
A sector of a circle is called a major if the major arc of the circle is a part of its boundary.
Shaded part of POQ is the major sector.

Question 5.
Write the formula to find the length of arc.
Solution:
Length of arc l = $$\frac{\mathrm{x}^{\circ}}{360^{\circ}}$$ 2πr (or) $$\frac{\mathrm{x}}{180^{\circ}}$$ πr.

Question 6.
Write the formula to find the area of a sector.
Solution:
Area of sector A = $$\frac{\mathrm{x}^{\circ}}{360^{\circ}}$$ πr2.

Question 7.
A pendulum swings through an angle of 30° and describe an arc 8.8 cm in length. Find the length of the pendulum.
Solution:
Given angle θ = 30°, l = 8.8 cm, r = ?
Length of an are

r = 0.8 × 7 × 3 = 16.8 cm

Question 8.
The radius of a circle is same as the side of a square. Find the ratio of their perimetrers.
Solution:
r = s
Perimeter of circle: Perimeter of square
2πr : 4r
πr : 2r
π : 2

Question 9.
Find the length of the arc of a circle of radius 14 cm which subtends an angle of 60° at the centre of the circle.
Solution:
θ = 60°, r = 14 cm
l = $$\frac{\theta}{360}$$ × 2πr
= $$\frac{60}{360}$$ × 2 × $$\frac{22}{7}$$ × 14
$$\frac{1}{6}$$ × 2 × 22 × 2
= $$\frac{22 \times 2}{3}$$ = $$\frac{44}{3}$$

Question 10.
If the radius of a semi-circular protractor is 7 cm, then find perimeter.
Solution:

r = 7 cm
p = $$\frac{36}{7}$$ r = $$\frac{36}{7}$$ × 7 = 36 cm

Question 11.
Find the area of the sector of a circle of radius 6 cm whose centred angle is 30°. (Take π = 3.14)
Solution:
r = 6 cm
θ = 30°
Area of sector = $$\frac{\theta}{360}$$ × πr2
= $$\frac{30}{360}$$ $$\frac{22}{7}$$ × 6 × 6
= $$\frac{1}{12}$$ × $$\frac{22}{7}$$ × 36
= $$\frac{11 \times 36}{6 \times 7}$$
= $$\frac{11 \times 6}{7}$$
= $$\frac{66}{7}$$ cm2

Question 12.
Find the area of a quadrant of a circle where the circumference of circle is 176 m.
Solution:
θ = 90°
Circumference = 176 m
2πr = 176
2 × $$\frac{22}{7}$$ × r = 176
r = 176 × $$\frac{7}{22}$$ × $$\frac{1}{2}$$
r = $$\frac{8 \times 7}{2}$$
r = 28 cm
A = $$\frac{\theta}{360}$$ × πr2
= $$\frac{90}{360}$$ × $$\frac{22}{7}$$ × 28 × 28
= $$\frac{1}{4}$$ × 22 × 4 × 28
= 11 × 2 × 28
A = 616 cm2

Question 13.
The minute hand of clock is 84 cm long. Find the distance covered by the tip of minute hand from 10 : 10 am to 10 : 25 am.
Solution:
60 min → 360
1 min → 6°
10 : 10 to 10 : 25 AM mean 15 min
15 min = 15 × 6° = 90°
θ = 90°
l = $$\frac{\theta}{360}$$ × 2πr
= $$\frac{90}{360}$$ × 2 × $$\frac{22}{7}$$ × 84
$$\frac{1}{4}$$ × 2 × $$\frac{22}{7}$$ × 84
= 11 × 12
l = 132 cm

Question 14.
The diameter of a car wheel is 42 cm. Find the number of complete revolutions it will make in moving 132 km.
Solution:
d = 42 cm
r = 21 cm
Circumference = 2πr
= 2 × $$\frac{22}{7}$$ × 21
= 44 × 3
= 132 cm
No. of revolutions
= $$\frac{132 \times 10^4}{132}$$ = 104 (∵ 132 km = 1320000 = 132 × 104)

Question 15.
Find the area of a square that can be inscribed in a circle of area $$\frac{1408}{7}$$ cm2.
Solution:
πr2 = $$\frac{1408}{7}$$
$$\frac{22}{7}$$ × r2 = $$\frac{1408}{7}$$
r2 = $$\frac{1408}{22}$$
r2 = $$\frac{704}{11}$$
r2 = 64
r = 8 cm

Question 16.
A circular arc of length 22 cm subtends an angle θ at the centre of the circle 21 cm. Find the value of ‘θ’.

Solution:
l = 22 cm
r = 21 cm
l = $$\frac{\theta}{360}$$ × 2πr
22 × $$\frac{\theta}{360}$$ × 2 × $$\frac{22}{7}$$ × 21
1 = $$\frac{\theta}{180}$$ × 3
$$\frac{\theta}{60}$$ = 1
θ = 60°

Question 17.
Find the perimeter of the sector of a circle of a radius 14 cm and central angle 45°.

Solution:
r = 14 cm
θ = 45°
Perimeter = l + 2r
= $$\frac{\theta}{360}$$ × 2πr + 2r
= $$\frac{45}{360}$$ × 2 × $$\frac{22}{7}$$ × 14 + 2 × 14
= $$\frac{1}{8}$$ × 2 × 22 × 2 + 28
= 11 + 28
= 39 cm

Question 18.
Two concentric circles are at O. Find the area of shaded region, if outer and inner radii are 14 cm and 7 cm respectively.

Solution:
R = 14 m
r = 7 m
Area of shaded region = π(R2 – r2)
= $$\frac{22}{7}$$(142 – 72)
= $$\frac{22}{7}$$(196 – 49)
= $$\frac{22}{7}$$ × 147
= 22 × 21
= 462 cm2

Question 19.
The circumference of two circles are in the ratio 4 : 5. What is the ratio of their radii?
Solution:
2πr1 : 2πr2 = 4 : 5
r1 : r2 = 4 : 5

Question 20.
If the perimeter of a circle is equal to that of a square, then find the ratio of their areas.
Solution:
4s = 2πr
⇒ s = $$\frac{\pi r}{2}$$ ⇒ s2 = πr2
⇒ ($$\frac{\pi r}{2}$$)2 : πr2 ⇒ $$\frac{\pi^2 r^2}{4}$$ : πr2
⇒ $$\frac{\pi}{4}$$ : 1 [π = $$\frac{22}{7}$$] ⇒ 14 : 11

Question 21.
If the radii of two circles are in the ratio of 4:3, then find the ratio of their areas.
Solution:
r1 : r2 = 4 : 3
⇒ πr12 : πr22
⇒ π 42 : π 32
⇒ 16 : 9

Question 22.
Find the area of a quadrant of a circle of radius 7 cm.
Solution:
r = 7
Area of quadrant = $$\frac{\pi r^2}{4}$$
= $$\frac{22 \times 7 \times 7}{7 \times 4}$$
= $$\frac{11 \times 7}{2}$$
= $$\frac{77}{2}$$ cm

Question 23.
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. (Take π = $$\frac{22}{7}$$)
Solution:
r = 6 cm
θ = 60°
Area of sector = $$\frac{\theta}{360}$$ × πr2
= $$\frac{60}{360}$$ × $$\frac{22}{7}$$ × 6 × 6 cm2
= $$\frac{132}{7}$$ cm2

Question 24.
A horse tied to a pole with 28m long rope. Find the perimeter where the horse can graze. (Take π = $$\frac{22}{7}$$)
Solution:

r = 28 cm
θ = 90°
Perimeter = 2πr
= 2 × $$\frac{22}{7}$$ × 28 = 44 × 4 = 176 cm

Question 25.
In a circle of diameter 42 cm, if an arc subtends an angle of 60° at the centre where π = $$\frac{22}{7}$$ then find the length of the arc.
Solution:
θ = 60°
d = 42 cm
r = 21 cm
l = $$\frac{\theta}{360}$$ × 2πr
= $$\frac{60}{360}$$ × 2 × $$\frac{22}{7}$$ × 21
= $$\frac{1}{6}$$ × 2 × 22 × 3 = 22 cm.

Question 26.
Find the perimeter of a semicircular protractor whose radius is ‘r’.
Solution:
P = πr + 2r

Question 27.
If the circumference of a circle increases from 2π it to 4π then its area ………………… the original area.
Solution:

C = 2πr
If r = 1
C = 2π
A = π(r)2 = π(1)2 = π
2πr = 4π
r = 2
New area = πr2 = π(2)2
= 4π = 4 × old area

Question 28.
If the difference between the circumference and the radius of a circle is 37cm, π = $$\frac{22}{7}$$, find the circumference (in cm) of the circle.
Solution:
2πr – r = 37 cm
(2 × $$\frac{22}{7}$$ – 1) r = 37
($$\frac{44 – 7}{7}$$)r = 37
$$\frac{37}{7}$$ r = 37
r = 7
C = 2πr
= 2 × $$\frac{22}{7}$$ × 7
= 44 cm

Question 29.
Area of a sector of a circle is $$\frac{1}{6}$$ to the area of circle. Find the degree measure of its minor arc.
Solution:
$$\frac{\theta}{360}$$ × πr2 = $$\frac{1}{6}$$ × πr2
$$\frac{\theta}{360}$$ = $$\frac{1}{6}$$
θ = 60°

Question 30.
Figure is a sector of circle of radius 10.5 cm, find the perimeter of the sector. (Take π = $$\frac{22}{7}$$
Solution:

r = 10.5 cm
θ = 60°
P = l + 2r
= $$\frac{\theta}{360}$$ × 2πr + 2r
= $$\frac{60}{360}$$ × 2 × $$\frac{22}{7}$$ × 10.5 + 2(10.5)
= $$\frac{1}{6}$$ × 2 × 22 × 1.5 + 21
= 22 × 0.5 + 21
= 11 + 21
= 32 cm

Question 31.
Assertion (A) : Area of sector if l = 10 cm, r = 9 cm is 10.5 cm
Reason (F) : Area of sector = $$\frac{\theta}{360}$$ × πr2
A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) A is true, R is true, but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
D) A is false, R is true.

Question 32.
Assertion (A) : Area of circle with 1 cm radius is π cm2.
Reason (R) : Length of the arc of the sector (l = $$\frac{\theta}{180}$$ × 2πr
A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) A is true, R is true, but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
C) A is true, R is false.

Question 33.
Assertion (A): Rddius of a circle is = $$\frac{7}{\sqrt{\pi}}$$ cm then area is 49 sq. units.
Reason (R) : Area of circle = πr2.
A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) A is true, R is true, but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.

Question 34.
Assertion (A): If a chord of length equal to the radius in a triangle then the triangle formed with that chord is an equilateral triangle.
Reason (R) : Area of ring π(R2 – r2).
A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) A is true, R is true, but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
B) A is true, R is true, but R is not the correct explanation of A.

Question 35.
Observe the figure, match the column.

A) Area of segment AYB
B) Area of sector OAYB

i) $$\frac{441}{4}$$√3
ii) $$\frac{21}{4}$$(88 – 21√3)
iii) 462
iv) 21/2
Solution:
A – i, B – iii

Question 36.
Choose correct matching.

A – (i), B – (ii).

Question 37.
Draw a rough diagram of minor segment of a circle and shade it.

Question 38.
What do we call the part a and b in the below circle ?

Solution:
‘a’ is minor segment and ‘b’ is major segment.

### 10th Class Maths Areas Related to Circles 2 Marks Important Questions

Question 1.
Find the area of quadrant of a circle whose circumference is 22 cm.
Solution:
Circumference C = 2πr = 22 cm
= 2 × $$\frac{22}{7}$$ × r = 22
r = $$\frac{22 \times 7}{2 \times 22}$$ = $$\frac{7}{2}$$ cm
Area of a quadrant = $$\frac{1}{4}$$ πr2
= $$\frac{1}{4}$$ × $$\frac{22}{7}$$ × $$\frac{7}{2}$$ × $$\frac{7}{2}$$
Therefore, area of quadrant = $$\frac{77}{8}$$ = 9.625 cm2

Question 2.
The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius of the circle.
Solution:
Let the radius = r cm
Circumference = diameter + 16.8
2 × $$\frac{22}{7}$$ r = 2r + 16.8
44r = 7 (2r + 16.8) = 14r + 16.8 × 7
44r – 14r = 117.6
30r = 117.6
r = $$\frac{117.6}{30}$$ = 3.92 cm
Therefore, radius r = 3.92 cm.

Question 3.
A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.
Solution:
Let the radius of the circle r cm
Distance covered by wheel in 1 revolution = $$\frac{\text { Total distance moved }}{\text { Number of revolutions }}$$
= $$\frac{11}{5000}$$km = $$\frac{11 \times 1000 \times 100}{5000}$$
That is circumference of wheel = 220 cm
So, 2πr = 220
2 × $$\frac{22}{7}$$ r = 220
r = 220 × $$\frac{7}{2 \times 22}$$ = 35 cm
Diameter d = 2 × r = 2 × 35 = 70 cm

Question 4.
A wheel has diameter 84 cm. Find how many complete revolutions must it take to cover 792 m.
Solution:
r = $$\frac{\text { Diameter }}{2}$$ = $$\frac{84}{2}$$ = 42 cm
Circumference of the wheel
2πr = 2 × $$\frac{22}{7}$$ × 42 = 264 cm = 2.64 m
No. of revolutions made to 792 m = $$\frac{792}{2.64}$$ = 300

Question 5.
Find the perimeter of a quadrant of a circle of radius 14 cm.
Solution:

θ = 90°
r = 14 m
= l + 2r = $$\frac{\theta}{360}$$ × 21r + 2r
= $$\frac{90}{360}$$ × 2 × $$\frac{22}{7}$$ × 14 + 2 × 14
= $$\frac{1}{4}$$ × 2 × 22 × 2 + 28
= 22 + 28 = 50 cm

Question 6.
Find the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm.
Solution:
Given r1 = 24 cm, r2 = 7 cm
πR2 = πr12 + πr22
R2 = 242 + 72
R2 = 576 + 49
R2 = 625
R = 25 cm
Diameter, d = 2R = 50 cm

Question 7.
Find the area of circle whose circumference is 22.
Solution:
Circumference of circle = 22 m
2πr = 22
2 × $$\frac{22}{7}$$ × r = 22
$$\frac{2 r}{7}$$ = 1
r = $$\frac{7}{2}$$ m
Area of circle = πr2
= $$\frac{22}{7}$$ × $$\frac{7}{2}$$ × $$\frac{7}{2}$$ cm2
= $$\frac{77}{2}$$ cm2

Question 8.
Find the diameter of a circle whose area is equal to the sum of the areas of two circles of radii 40 cm and 9 cm.
Solution:
Given r1 = 40 cm, r2 = 9 cm
πR2 = πr12 + πr22
R2 = r12 + r22
R2 = 402 + 92
R2 = 1600 + 81
R2 = 1681
R = $$\sqrt{1681}$$
R = 41 cm
Diameter = 2 × 41 cm = 82 cm

Question 9.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of minor segment. (Use π = 3.14)
Solution:
θ = 90°

Area of right-angled triangle
= $$\frac{1}{2}$$ × 10 × 10 = 50 cm2
Area of minor segment = Area of sector – Area of right triangle
= 78.5 – 50 = 28.5 cm2

Question 10.
A piece of wire 22 cm long is bent into the form an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle.
Solution:
Given l = 22 cm
θ = 60°
l = $$\frac{\theta}{360}$$ × 2πr
22 = $$\frac{60}{360}$$ × 2 × $$\frac{22}{7}$$ × r
1 = $$\frac{1}{6}$$ × 2 × $$\frac{1}{7}$$ × r
1 = $$\frac{r}{21}$$
r = 21 m

Question 11.
An iron wire when bent in the form of a square encloses an arrea of 121 cm2. If the same wire is bent in the form of a circle, then find the circumference of circle.
Solution:
Area of square = 121 m2
S2 = 121
⇒ S = $$\sqrt{121}$$
S = 11 cm
⇒ 4S = 4 × 11 = 44 cm
2πr = 44
⇒ 2 × $$\frac{22}{7}$$ × r = 44
r = 7 cm
Area of circle = πr2
= $$\frac{22}{7}$$ × 7 × 7 cm2
= 154 cm2

Question 12.
If the radius of the circle is 6 cm and the length of an arc is 12 cm. Find the area of sector.
Solution:
Length of arc, l = 12 cm
Area of sector = $$\frac{\ell \mathrm{r}}{2}$$
= $$\frac{12 \times 6}{2}$$
= 6 × 6
= 36 cm2

Question 13.
PA is the tangent drawn to a circle of whose centre is ‘O’, OA is the radius and P is the external point of the circle. If PA = 24 cm, OP = 25 cm., then find its radius.
Solution:
Given PA = 24 cm, OP = 25 cm
We have OP2 = OA2 + PA2
⇒ (25)2 = OA2 + (24)2
⇒ 625 = OA2 + 576
⇒ OA2 = 49
⇒ OA = 7 cm

Question 14.
OABC is a Rhombus whose three vertices A, B and C lie on a circle with centre ‘O’. If the radius of the circle is 10 cm. Find the area of the Rhombus.
Solution:

Join OB
OA = OB (Sides of Rhombus)
∴ ΔOAB is an equilateral triangle.
Area of equilateral ΔOAB
= $$\frac{\sqrt{3}}{4}$$a2 = $$\frac{\sqrt{3}}{4}$$ × (10)2 = 25√3 cm2
∴ Area of Rhombus = 2 × area of ΔOAB
= 2 × 25√3 cm2
= 50√3 cm2

Question 15.
The minute hand of a clock is $$\sqrt{21}$$ cm long. Find the area of described by the minute hand on the face of the clock between 7 : 00 am to 7 : 05 am (π = $$\frac{\sqrt{22}}{7}$$)
Solution:
60 min → 360°
1 min → $$\frac{360^{\circ}}{60^{\circ}}$$ = 6°
Time between 7 am to 1 : 05 am = 5 min
Angle = 5 × 6°= 30°
Radius, r = $$\sqrt{21}$$ cm
Area swept = $$\frac{\theta}{360}$$ × πr2
= $$\frac{30^{\circ}}{360^{\circ}}$$ × $$\frac{22}{7}$$ × $$\sqrt{21}$$ × $$\sqrt{21}$$ = $$\frac{1}{12}$$ $$\frac{22}{7}$$ × 21
= $$\frac{11}{2}$$m2 = 5.5 m2

Question 16.
Find the area of quadrant of a circle, where the circumference of a circle is 44 cm. (π = $$\frac{\sqrt{22}}{7}$$)
Solution:
2πr = 44
2 × $$\frac{22}{7}$$ × r = 44
r = 7 cm,
θ = 90°
Area of quadrant = $$\frac{\theta}{360}$$ × πr2
= $$\frac{90^{\circ}}{360^{\circ}}$$ × $$\frac{22}{7}$$ × 7 × 7 cm2
= $$\frac{1}{4}$$ × 22 × 7cm2 = $$\frac{77}{2}$$ = 38.5 cm2

Question 17.
Find the area of a sector of a circle whose radius is 7 cm and angle at the centre is 60°.
Solution:
Radius = 7 cm, Angle at centre = 60°
Area of the sector = $$\frac{x}{360}$$ × πr2
= $$\frac{60 \times \frac{22}{7} \times 7 \times 7}{360}$$ = $$\frac{154}{6}$$
= $$\frac{60 \times \frac{22}{7} \times 7 \times 7}{360}$$ = $$\frac{154}{6}$$ = 25.66 cm

Question 18.
Find the area of a sector whose radius is 10 cm. and which make right angle at the center. (Take π = 3.14).
Solution:
Radius of the sector = 10 cm
Angle of sector = x° = 90°
Area of sector A = $$\frac{x^{\circ}}{360^{\circ}}$$ × πr2
= $$\frac{90}{360}$$ × 3.14 × (10)2
= $$\frac{1}{4}$$ × 3.14 × 100
= $$\frac{1}{4}$$ × 3.14 = 78.5 cm2
(OR)
Sector angle = right angle
So, the Area of quadrant sector = $$\frac{\pi r^2}{4}$$
= $$\frac{3.14 \times(10)^2}{4}$$ = $$\frac{3.14 \times 100}{4}$$
= $$\frac{314}{4}$$ = 78.5 cm2

Question 19.
Find the area of a sector whose radius is 10 cm. and which make right angle at the center. (Take π = 3.14).
Solution:
Radius of the sector =10 cm
Angle of sector = x° = 90°
Area of sector = A = $$\frac{x^{\circ}}{360^{\circ}}$$ × πr2
= $$\frac{90}{360}$$ × 3.14 × (10)2
= $$\frac{1}{4}$$ × 3.14 × 100
= $$\frac{1}{4}$$ × 314 = 78.5 cm2

Question 20.
Find the area of the shaded part in the given figure.

Solution:
Area of shaded part = Area of semi-circle – Area of triangle
= $$\frac{\pi r^2}{2}$$ – $$\frac{1}{2}$$bh
= $$\frac{\frac{22}{7} \times 7 \times 7}{2}$$ – $$\frac{1}{2}$$ × 14 × 6
= 11 × 7 – 7 × 6
= 77 – 42 = 35 sq. cm.

### 10th Class Maths Areas Related to Circles 4 Marks Important Questions

Question 1.
A horse is tied to a pole with 28 m long string. Find the area where the horse can graze.
Solution:
Given horse can graze the circular and radius r = 28 m
Area of land grazed by horse
= πr2 = $$\frac{22}{7}$$ × 28 × 28 = 2464 m2

Question 2.
The circumferences of two circle s are in the ratio 2 : 3. Find their areas.
Solution:
Let radius of circules are r1 : r2
Ratio of their circumferences
= 2πr1 : 2πr2 = 2 : 3
= r1 : r2 = 2 : 3
Areas = πr12 : 2πr22
= $$\frac{22}{7}$$ × 22 and $$\frac{22}{7}$$ × 32
= $$\frac{88}{7}$$ m2 and $$\frac{198}{7}$$ m2

Question 3.
If a square is inscribed in a circle. Find the ratio of the areas the circle and the square.
Solution:
Let side of square = x units
area of square = x.x = x2 sq. units.
diagonal of square = √2.Side = √2.x
radius r = $$\frac{\text { Length of diagonal }}{2}$$
= $$\frac{\sqrt{2} \cdot x}{2}$$ units
Area of circle = πr2
= π.($$\frac{\sqrt{2} \cdot x}{2}$$)2
= π.$$\frac{2 \cdot x}{4}$$
= $$\frac{\pi x^2}{2}$$
Area of circle : Area of square
= $$\frac{\pi x^2}{2}$$ : x2 = $$\frac{\pi}{2}$$ : 1
Therefore, ratio = π : 2

Question 4.
A sector is cut from a circle of radius 21 cm, the angle of the sector is 150°. Find the length of its arc and area.
Solution:
Given radius r = 21 cm and θ = 150°,
Length of arc

Question 5.
Find the angle subtended at the centre of a circle of radius 5 cm by an arc of length $$\frac{5 \pi}{3}$$ cm.
Solution:
Radius of circle r = 5 cm
Let angle = θ

Therefore, angle subtended at the centre θ = 60°

Question 6.
A drain cover is made from a square metal plate of side 40 cm having 441 holes of diameter 1 m each drilled in it. Find the area of the remaining square plate.
Solution:
Side of square S = 40 cm, radius = $$\frac{d}{2}$$ = $$\frac{1}{2}$$ cm
Area of square = S.S = 40 × 40 = 1600 cm2
Area of each hole
= πr2 = $$\frac{22}{7}$$($$\frac{1}{2}$$)2 = $$\frac{11}{14}$$ cm2
Area of 441 holes
= 441 × $$\frac{11}{14}$$ = 346.5 cm2
Area of the remaining plate = 1600 – 346.5 = 1253.5 cm2

Question 7.
The area of a circular play ground is 22176 cm2. Find the cost of fencing this ground at the rate of ₹ 50 per metre.
Solution:
πr2 = 22176
$$\frac{22}{7}$$ × r2 = 22176
r2 = 22176 × $$\frac{7}{22}$$
r2 = 1008 × 7
r2 = 7056
r = 84 m
2πr = 2 × $$\frac{22}{7}$$ × 84
= 44 × 12
= 528 cm
c = $$\frac{528}{100}$$ cm
c = 5.28 m
Cost of fencing per metre = ₹ 50
Total cost = 50 × 5.28 = ₹ 264

Question 8.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the area of the sector formed by the arc. Also, find the length of the arc.
Solution:
r = 21 cm
θ = 60°
Area of sector = $$\frac{\theta}{360}$$ × πr2
= $$\frac{60}{360}$$ × $$\frac{22}{7}$$ × 21 × 21 = $$\frac{1}{6}$$ × 22 × 3 × 21
= 11 × 21 = 231 = cm2
Lenth of arc of sector = $$\frac{\theta}{360}$$ × 2πr
= $$\frac{60^{\circ}}{360^{\circ}}$$ × 2 × $$\frac{22}{7}$$ × 21 = $$\frac{1}{6}$$ × 2 × 22 × 3 = 22 cm

Question 9.
Find the area of the sector of a circle of radius 6 cm and of central angle 90°. Also, find the area of corresponding major sector.
Solution:
r = 7 cm
θ = 90°
area of sector = $$\frac{\theta}{360}$$ × πr2
= $$\frac{90^{\circ}}{360^{\circ}}$$ × $$\frac{22}{7}$$ × 7 × 7 cm2
= $$\frac{1}{4}$$ × 22 × 7 cm2
= $$\frac{77}{2}$$ cm2 = 38.5 cm2
Area of major sector = Area of circle – area of minor sector
= πr2 – 38.5 cm2
= $$\frac{22}{7}$$ × 7 × 7 cm2 – 38.5cm2
= 154 – 38.5 cm2
= 119.5 cm2

Question 10.
Find the area of the minor segment of a circle of radius 42 cm, if length of the corresponding arc is 44 cm.
Solution:

r = 42 cm
l = 44 cm
44 = $$\frac{\theta}{360}$$ × 2πr
44 = $$\frac{\theta}{360}$$ × 2 × $$\frac{22}{7}$$ × 42
1 = $$\frac{\theta}{360}$$ × 6
θ = 60°
ΔOAB is an equilateral triangle.
Area of ΔOAB = $$\frac{\sqrt{3}}{4}$$a2
= $$\frac{\sqrt{3}}{4}$$ × 42 × 42
= $$\frac{\sqrt{3}}{2}$$ × 21 × 42
= 21 × 21 √3
= 441√3 m2
Area of sector = $$\frac{\theta}{360}$$ × πr2
= $$\frac{60^{\circ}}{360^{\circ}}$$ × $$\frac{22}{7}$$ × 42 × 42
= $$\frac{1}{6}$$ × 22 × 6 × 42
= 22 × 42
= 924 cm2
Area of minor segment = Area of sector – area of triangle
= (924 – 441√3 ) cm2

Question 11.
In the given figure, two concentric circles with centre O are shown. Radii of the circles are 2 cm and 5 cm respectively. Find the area of the shaded region.
Solution:

Given θ = 60°
r1 = 2 cm
r1 = 5 cm
Area of big sector = $$\frac{\theta}{360}$$ × πr22
= $$\frac{60^{\circ}}{360^{\circ}}$$ × $$\frac{22}{7}$$ × 5 × 5
= $$\frac{1}{6}$$ × $$\frac{22}{7}$$ × 25
= $$\frac{11 \times 25}{21}$$ = $$\frac{275}{21}$$ cm2
Area of small sector = $$\frac{60^{\circ}}{360^{\circ}}$$ × πr22
= $$\frac{1}{6}$$ × $$\frac{22}{7}$$ × 22
= $$\frac{1}{6}$$ × $$\frac{22}{7}$$ × 4 = $$\frac{44}{21}$$
Area of shaded region = $$\frac{275}{21}$$ – $$\frac{44}{21}$$
= $$\frac{231}{21}$$ cm2

Question 12.
In the given figure, ‘O’ is the centre of circle such that diameter AB =13 cm and AC =12 cm. BC is joined. Find the area of shaded region (use = 3.14)

Solution:
‘O’ is the centre of circle
AB is the diameter.
AB = 13 cm
AC = 12 cm
∠ACB = 90° (Angle is a semi circle)
∴ ΔACB is a right angled triangle.
AB2 = AC2 + BC2
132 = 122 + BC2
169 – 144 = BC2
BC2 = 25
BC = $$\sqrt{25}$$
BC = 5 cm
Ar ΔACB = $$\frac{1}{2}$$ × BC × AC
= $$\frac{1}{2}$$ × 5 × 12 cm2
= 30 cm2
r = $$\frac{13}{2}$$ cm
Area of semi circle
= $$\frac{\pi r^2}{2}$$ = $$\frac{3.14 \times \frac{13}{2} \times \frac{13}{2} \mathrm{~cm}^2}{2}$$
= $$\frac{3.14 \times 169}{8}$$
= $$\frac{530.66}{8}$$
= 66.33 cm2
Area of shaded region = 66.33 – 30 cm2 = 36.33 cm2

Question 13.
In the figure, sectors of two concentric circles of radii 7 cm and 3.5 cm are given. Find the area of shaded region. (π = $$\frac{22}{7}$$)

Solution:
r1 = 3.5, r2 = 7 cm
Area of sector (big) = $$\frac{\theta}{360}$$ × πr2
= $$\frac{30}{360}$$ × $$\frac{22}{7}$$ × 7 × 7 cm2 = $$\frac{154}{2}$$ = $$\frac{77}{6}$$ cm2
Area of sector (smaller) = $$\frac{\theta}{360}$$ × πr12
= $$\frac{30^{\circ}}{360^{\circ}}$$ × $$\frac{22}{7}$$ × 3.5 × 3.5
= $$\frac{1}{12}$$ × $$\frac{22}{7}$$ × $$\frac{7}{2}$$ × $$\frac{7}{2}$$ = $$\frac{77}{24}$$ cm2
Area of shaded region = $$\frac{77}{6}$$ – $$\frac{77}{24}$$ = $$\frac{308}{24}$$ – $$\frac{77}{24}$$ = $$\frac{231}{24}$$ cm2

Question 14.
Find the area of the corresponding major sector of a circle of radius 28 cm and the central angle is 45°.
Solution:
r = 28 cm, θ = 45°
Area of minor sector = $$\frac{\theta}{360^{\circ}}$$ × πr2
= $$\frac{45^{\circ}}{360^{\circ}}$$ × $$\frac{22}{7}$$ × 28 × 28 cm2
= $$\frac{1}{8}$$ × 22 × 4 × 28 cm2 = 28 × 11 cm2 = 616 cm2
Area of major sector= Area of circle – Area of minor sector
= πr2 – 616 cm2
= $$\frac{22}{7}$$ × 28 × 28 – 28 × 11 cm2
= 28[$$\frac{22}{7}$$ × 28 – 11] = 28[88 – 11]
= 28 × 77 = 2156 cm2

Question 15.
In the figure AOB is a sector of angle 60° of a circle with centre ‘O’ and radius 17 cm. If AP⊥OB and AP = 15 cm then find the area of shaded region.

Solution:
θ = 60°
r = 17 cm
Area of sector = $$\frac{\theta}{360^{\circ}}$$ × πr2
= $$\frac{60^{\circ}}{360^{\circ}}$$ × $$\frac{22}{7}$$ × 17 × 17 = $$\frac{1}{6}$$ × $$\frac{22}{7}$$ × 289 cm2
$$\frac{11 \times 289}{21}$$ = $$\frac{3179}{21}$$ cm2
Area of ΔOPA:

AP⊥DB
∠APO = 90°
∠AOP = 60°
∠PAO = 180° – 90° – 60°
= 180 – 150° = 30°
cos 60° = $$\frac{P O}{17}$$
$$\frac{1}{2}$$ = $$\frac{P O}{17}$$
PO = $$\frac{17}{2}$$ = 8.5
sin 60° = $$\frac{A P}{17}$$
$$\frac{\sqrt{3}}{2}$$ = $$\frac{A P}{17}$$
AP = 8.5√3 cm
Area of ΔOAP = $$\frac{1}{2}$$ × OP × AP
= $$\frac{1}{2}$$ × 8.5 × 8.573 = $$\frac{72.25 \sqrt{3}}{2}$$
Area of shaded region = Area of sector – Ar ΔOAP
= $$\frac{3179}{2}$$ – $$\frac{72.25 \sqrt{3}}{2}$$ cm2

Question 16.
A wire when bent in the form of an equilateral triangle encloses an area of 121√3 cm2. If the wire is bent in the form of a circle find the area enclosed by the circle. (Use π = $$\frac{22}{7}$$)
Solution:
Area of equilateral triangle = 121√3 cm2
$$\frac{\sqrt{3}}{4}$$a2 =121√3
a2 = 121 × 4
a = 11 × 2
a = 22 cm
Perimeter = 3a = 66 cm
2πr = 66 cm
2 × $$\frac{22}{7}$$ × r = 66 ⇒ 2 × $$\frac{1}{7}$$ × r = 3
r = $$\frac{21}{2}$$ cm
Area of circle = πr2
= $$\frac{22}{7}$$ × $$\frac{21}{2}$$ × $$\frac{21}{2}$$ cm2 11 × 3 × $$\frac{21}{2}$$ cm2
= $$\frac{33 \times 21}{2}$$ cm2 = 33 × 10.5 cm2 = 346.5 cm2

Question 17.
Find the area of shaded region in the figure, if BC = BD = 8 cm, AC = AD = 15 cm and ‘O’ is the centre of the circle (Take π = 3.14)

Solution:
‘O’ is centre of circle
AC = AD = 15 cm
BC = BD = 8 cm
ΔACB and ΔADB are right angled triangles
∴ ∠ACB = 90° = ∠ADB
AB2 = AC2 + BC2
AB2 = 152 + 82
AB2 = 225 + 64
AB2 = 289
AB = $$\sqrt{289}$$
AB = 17 cm
r = $$\frac{17}{2}$$ cm
ar ΔABC = $$\frac{1}{2}$$ × BC × AD = $$\frac{1}{2}$$ × 8 × 15 = 60 cm2
∴ Area of two triangles = (60 + 60) cm2 = 120 cm2
Area of shaded region = Area of circle – Area of both triangles
= πr2 – 120 cm2
= 3.14 × $$\frac{17}{2}$$ × $$\frac{17}{2}$$ – 120 cm2
= 3.14 × 8.5 × 8.5 – 120 cm2
= 226.865 – 120 cm2
= 106.865 cm2

Question 18.
In the given figure, a semi circle of radius 7 cm is inscribed in a rectangle. Find the area of shaded region.

Solution:
In a semi circle r = 7 cm
Area = $$\frac{\pi r^2}{2}$$ = $$\frac{22}{7}$$ × $$\frac{7 \times 7}{2}$$
= 11 × 7 = 77 cm2
In a rectangle,
l = 7 + 7 = 14 cm
b = 7 cm
Area = 7 × 14 = 98 cm2
Area of shaded region = 98 – 77 cm2 = 21 cm2

Question 19.
Find the area of the shaded region in the given figure.
ABCD is a square of side 10.5 cm.

Solution:
Side of the Square = 10.5 cm
Radius of the circle (r) = $$\frac{10.5}{2}$$

Area of the square = (Side)2 = (10.5)2 = 110.25 cm2
Area of the circle = pr2
= $$\frac{22}{7}$$ × $$\frac{10.5}{2}$$ × $$\frac{10.5}{2}$$ = 86.625 cm2
Area of the shaded region = Area of the square – Area of the circle
= 110.25 – 86.625
= 23.625cm2

Question 20.
A chord of a circle of radius 10 cm. subtends a right angle at the centre. Find the area of the corresponding minor segment (use p = 3.14).
Solution:

Radius of circle (r) = 10 cm
Sector angle (x) = 90°
Radius of sector (r) = 10 cm.
Area of sector OACB = $$\frac{x}{360}$$ × πr2
= $$\frac{90^{\circ}}{360^{\circ}}$$ × 3.14 × 10 × 10
= 78.5 Sq. cm
Area of D AOB = $$\frac{1}{2}$$ × 10 × 10
= 50 cm2
Area of the Minor Segment = Area of sector OACB – Area of D OAB
= 78.5 – 50.0 = 28.5 cm2

### 10th Class Maths Areas Related to Circles 8 Marks Important Questions

Question 1.
Find the area of the shaded region of the given figure, where a circular arc of radius 6 cm has been drawn with vertex of an equilateral triangle OAB of side 12 cm as centre.
Solution:
Given radius of the circle r = 6 cm
Side of triangle = 12 cm
Angle in the minor sector (or) angle in the equailateral triangle = 60°
Angle in the major sector θ = 360° – 60° = 300°
Area of the shaded region = (Area of equailateral triangle + Area of major sector)
= ($$\frac{\sqrt{3}}{4}$$a2 + $$\frac{\theta}{360}$$ πr2)
= $$\frac{\sqrt{3}}{4}$$ × 12 × 12 + $$\frac{300}{360}$$ × 3.14 × 6 × 6
= (36√3 + 30 × 3.14)
= 114.235 + 92.2
= 208.435 cm2
Therefore, area of the shaded region
= 208.435 cm2

Question 2.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find
i) Length of the arc
ii) Area of the sector formed by the arc.
iii) Area of the segment formed by the corresponding chord of the arc.
iv) Find the area of the major segment.
Solution:
Given radius of the circle = 21 cm
Angle subtended by the arc θ = 60°

= 231 – 19.953
Area of minor segment = 40.047 cm2

iv) Area of major segment
= Area of circle – Area of minor segment
= πr2 – 40.047
= $$\frac{22}{7}$$ × 21 × 21 – 40.047
= 1386 – 40.047
Therefore, area of major segment
= 1345.953 cm2

Question 3.
The length of minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
Given length of clock (or) radius of circle r = 14 cm
Angle made by the minute hand in 5 minutes = θ = 30°
Area of the sector

Area swept by the minutes hand in 5 minutes = $$\frac{154}{3}$$ cm2

Question 4.
Find the area of the sector of a circle with radius 4 cm and of an angle 30°. Also find the area of the corresponding major sector.
Solution:
Given radius of the circle r = 4 cm
Angle of the sector θ = 30°
Area of the sector (OAXB)

Therefore area of the sector = 4.153 cm2
Area of the corresponding major sector
= Area of sector OAYB
= Area of the circle – Area of OAXB
= πr2 – $$\frac{x}{360}$$πr2 = (1 – $$\frac{x}{360}$$)πr2
= (1 – $$\frac{30}{360}$$) × $$\frac{22}{7}$$ × 4 × 4
= ($$\frac{1}{1}$$ – $$\frac{1}{12}$$) × 3.14 × 4 × 4
= ($$\frac{12 -1}{12}$$) × 3.14 × 4 × 4
= $$\frac{11}{12}$$ × 3.14 × 4 × 4 = 46.05 cm2
Therefore, area of the corresponding major sector = 46.05 cm2

Question 5.
In a circle with centre O and radius 5 cm, AB is a chord of length 5√3 cm.
Find the area of the sector AOB.
Solution:
OA = OB = r = 5 cm

Length of the chord AB = 5√3 cm
Draw OM⊥AB, OM bisects AB
That is AM = BM = $$\frac{5 \sqrt{3}}{2}$$
In ΔAOM, ∠OMA = 90°, ∠AOM = θ
sin θ $$\frac{\mathrm{AM}}{\mathrm{OA}}$$ = $$\frac{\frac{5 \sqrt{3}}{2}}{5}$$ = $$\frac{\sqrt{3}}{2}$$ = sin 60°
Therefore θ = 60°
That is ∠AOB = 2 × 60 = 120
Area of sector OAXB
= $$\frac{\angle \mathrm{AOB}}{360}$$ × πr2 = $$\frac{120}{360}$$ × 3.14 × 5 × 5
Area of sector OAXB = $$\frac{3.14}{3}$$ × 5 × 5
= $$\frac{78.5}{3}$$ = 26.17 cm2

Question 6.
An arc of length 20π cm subtends an angle of 144° at the centre of a circle. Find the radius of the circle.
Solution:
Given length of the arc = 20π cm,
Angle subtended by arc θ = 144°
Length of the arc l = $$\frac{\theta}{360}$$ × 2πr
= $$\frac{144}{360}$$ × 2πr = 20π
r = $$\frac{20 \pi \times 360}{144 \times 2 \pi}$$
Therefore, radius of circle r = 25 cm

Question 7.
The minute hand of a clock is $$\sqrt{21}$$ cm long. Find the area described by the minute hand on the face of the clock between 7.00 AM and 7.05 AM.
Solution:
Given the length of minute hand = radius of circle r = $$\sqrt{21}$$ cm

= $$\frac{\theta}{360}$$ × πr2
= $$\frac{30}{360}$$ × $$\frac{22}{7}$$ × $$\sqrt{21}$$ × $$\sqrt{21}$$
= $$\frac{30}{360}$$ × $$\frac{22}{7}$$ × 21
= $$\frac{11}{2}$$ = 5.5 cm2
Area described by the minute hand in 5 minutes = 5.5 cm2

Question 8.
The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 8 AM and 8.25 AM.
Solution:
Given length of the minute hand = Radius of the circle (r) = 10 cm
Minute hand makes the angle by moving 25 minutes at the centre θ = 150°
Area described by the minute hand

Area described by the minutes hand in 25 minutes = 130.95 cm2

Question 9.
A horse is placed for grazing inside a rectangular field 70 m by 52 m and is togethered to one corner by a rope 21 m long. On how much area can it graze ?
Solution:

Given
Radius of the sector r = 21 m
Angle of the sector θ = 90°
Area of the sector = $$\frac{\theta}{360}$$ × πr2
= $$\frac{90}{360}$$ × $$\frac{22}{7}$$ × 21 × 21
= $$\frac{11 \times 21 \times 3}{2}$$
= 346.5 cm2
Therefore, area to be grazed by the horse = 346.5 cm2

Question 10.
ABCD is a flower bed. If OA = 21 m and OC = 14 m, find the area of the bed.
Solution:
Given the radius of large sector (or) OA = OB = 21 m A

Radius of the small sector (r) OC = OD = 14 cm
Angle of the sector = ∠AOB = ∠COD = 90°
Area of the flower bed = Area of large sector – Area of small sector
= $$\frac{\theta}{360}$$ × πR2 – $$\frac{\theta}{360}$$ × πr2
= $$\frac{\theta}{360}$$ × π(R2 – r2)
= $$\frac{90}{360}$$ × $$\frac{22}{7}$$ (212 – 142)
= $$\frac{1}{4}$$ × $$\frac{22}{7}$$ × (21 + 14) (21 – 14)
= $$\frac{1}{4}$$ × $$\frac{22}{7}$$ × 35 × 7
= $$\frac{22 \times 35}{4}$$ = 192.5 m2
Therefore, area of the flower bed = 192.5 m2

Question 11.
ABCD is a quadrant of a circle of radius 14 cm with AC as diameter, a semi-circle is drawn. Find the area of the shaded portion.
Solution:

Given in ΔABC, ∠B = 90°
AB = BC = 14 m
AC2 = AB2 + BC2
AC = $$\sqrt{14^2+14^2}$$
= $$\sqrt{14^2 \times 2}$$
AC = 14√2cm
Diameter of semi circle AC = 14√2 Radius of semi circle
= $$\frac{A C}{2}$$ = $$\frac{14 \sqrt{2}}{2}$$ = 7√2 cm
= Area of APCQA
= Area of ACQA – Area of ACPA
= Area of semi circle – (Area of ABCPA – Area of ΔABC)

Area of shaded portion = 98 cm2

Question 12.
On a circular table cover of radius 32 cm a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design.

Solution:
Given ΔABC is an equilateral triangle.
That is ∠A = ∠B = ∠C = 60°
and AB = BC = AC
Radius of the circle OA = OB = OC = 32 cm
Each side of ΔABC can make the angle at the centre is 120°
So, θ = ∠BOC = 120°
Draw OD⊥BC
That is OD bisects BC and ∠BOC
BD = CD = $$\frac{B C}{2}$$
∠BOD = ∠COD = $$\frac{1}{2}$$ ∠BOC
= $$\frac{1}{2}$$ × 120 = 60°
In ΔOBD, ∠ODB = 90°

BC = 2.BD = 2 × 16√3 = 32√3 cm
Area of shaded region = 3 (area of sector – area of ΔABC)

= 1888.11 cm2
Area of the shaded region = 1888.11 cm2

Question 13.
In the given figure O is the centre of the circle of radius 28 cm. Find the area of minor segment AXB.
Solution:
Given radius of the circle (r)
OA = OB = 28 cm

Angle at sector ∠AOB (θ) = 90°
Area of minor segment = Area of sector – Area of ΔAOB
= $$\frac{\theta}{360}$$πr2 – $$\frac{1}{2}$$ × AB × OX
In ΔAOB, ∠AOB = 90°,
Draw OX⊥AB
OX bisects AB and ∠AOB
that is AX = BX = $$\frac{A B}{2}$$
and ∠AOX = ∠BOX = $$\frac{\angle \mathrm{AOB}}{2}$$
∠AOX = ∠BOX = $$\frac{90}{2}$$ = 45°
In ΔAOX, ∠AOX = 45°

= 22 × 28 – 14 × 14 × √2 × √2
= 616 – 392 = 224 cm2
Therefore, area of minor segment = 224 cm2

Question 14.
From a circular piece of cardboard of radius 3 cm two sectors of 90° have been cutoff. Find the perimeter of the remaining portion nearest hundredth
centimeters (π = $$\frac{22}{7}$$).
Solution:
Given radius of the circle r = 3 cm

Angle of each sector θ = 90°
two quadrant circles = semi circle
So, remaining portion is a semi circle
Perimeter of semi circle = πr
= $$\frac{22}{7}$$ × 3 $$\frac{22}{7}$$ = 9.428 cm

Question 15.
A square OABC is inscribed in a quad-rant OPBQ of a circle. If OA = 21 cm find the area of the shaded region.
Solution:
Given OA = AB = BC = OC = 21 cm

OB2 = OA2 + AB2
OB2 = 212 + 212 = 2 × 212
OB = $$\sqrt{2 \times 21^2}$$ = 21√2 cm
OP = OB = OQ = 21√2 cm
Area of the shaded region = Area of OPBQ – Area of OABC
= $$\frac{1}{4}$$πr2 – OA × AB
= $$\frac{1}{4}$$ × $$\frac{22}{7}$$ × 21√2 × 21√2 – 21 × 21
= 11 × 3 × 21 – 21 × 21
= 693 – 441 = 252 cm2
Therefore, area of shaded region = 252 cm2

Question 16.
A chord of circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
i) Minor segment
ii) Major segment (use p = 3. 14)
Solution:
Radius of circle (r) = 10 cm
Sector angle (x) = 90°
Radius of sector (r) = 10 cm.

Area of sector OACB = $$\frac{x}{360}$$ × πr2
= $$\frac{90^{\circ}}{360^{\circ}}$$ × 3.14 × 10 × 10
= 78.5 Sq. cm
Area of Δ AOB = $$\frac{1}{2}$$ × 10 × 10= 50 cm2
Area of the Minor Segment
= Area of sector OACB – Area of Δ OAB
= 78.5 – 50.0 = 28.5 cm2
Area of Major Segment
= Area of the circle – Area of the Minor Segment
= (3.14 × 10 × 10) – 28.5
= 314 – 28.5 = 285.5 cm2

Question 17.
Find the area of the segment shaded in the figure in which PQ = 12 cm, PR = 5 cm and QR is the diameter of the circle with centre ‘O’. (Take p = $$\frac{22}{7}$$

Solution:
To find the area of the segment shaded in the given figure.
Here PQ’ = 12 cm; PR’ = 5 cm; QR’ is diameter
Now PQOR is a semicircle
then angle in a semicircle is 90°
then ∠QPR = 90°
∴ ΔPQR is a right angled triangle
∴ Area of ΔPQR = $$\frac{1}{2}$$ bh
= $$\frac{1}{2}$$ × PQ × PR
= $$\frac{1}{2}$$ × 12 × 5 = 30 cm2 ______ (1)
Now the area of shaded part = area of semicircle – area of ΔPQR
= $$\frac{1}{2}$$πr2 – 30 cm2 _______ (2)
In ΔPQR, QR2 = PQ2 + PR2 (from Pythagoras theorem)
QR2 = 122 + 52
= 144 + 25 = 169 = 132
\QR = 13 then
Radius of the circle (r) = QO = $$\frac{Q R}{2}$$ = $$\frac{13}{2}$$ = 6.5 cm
then area of semicircle
= $$\frac{1}{2}$$ Pr2
= $$\frac{1}{2}$$ × $$\frac{22}{7}$$ × $$\frac{13}{2}$$ × $$\frac{13}{2}$$ = 66.39 cm2 ________ (3)
Now putting the values of (1) and (3) in (2) we get
Area of shaded part = (66.39 – 30)
= 36.39 cm2.

Question 18.
Ten identical mementos is made by a school to awarding 10 students for first prize winners in games. If each memento is made as shown in figure (shaded portion) its base PQRS is silver plated from the front side at the rate of 20 per square cm. Find the total cost of the silver plating of 10 mementos. (OR = 5cm, RQ = 6cm, PS = 8 cm)

Solution:
Area of right angled triangle
= $$\frac{1}{2}$$ × 13 × 11 = $$\frac{143}{2}$$ = 71.5 cm2
Area of the sector = $$\frac{90^{\circ}}{360^{\circ}}$$ × $$\frac{22}{7}$$ × 52
= $$\frac{1}{4}$$ × $$\frac{22}{7}$$ × 25
= 19.64 cm2
∴ Area of the shaded portion
= 71.5 – 19.64 = 51.86 cm2
The rate of silver plating per one square cm = Rs. 20
∴ The total cost of the plating of 10 mementos = 51.86 × 10 × 20
= Rs. 10,372

Question 19.
A square ODEF is inscribed in a quadrant OPEQ of circle and OD = 14√2 cm. Aarthi said that “the area of shaded region is 224 cm2.” Do you agree ? Give reasons.

Solution:
Side of the square = OD = 14√2 cm
Length of the diagonal = OE
= √2 (OD)
= √2 (14√2) = 28 cm
Radius of the quadrant OPEQ = r = diagonal of the square ODEF.
Area of the quadrant OPEQ = $$\frac{1}{4}$$ pr2
= $$\frac{1}{4}$$ × $$\frac{22}{7}$$ × 28 × 28
= 22 × 28
= 616 cm2
Area of the square ODEF = (OD)2
= (14√2)2 = 196 × 2 = 392 cm2
∴ The area of the shaded region
= Area of the quadrant OPEQ – Area of the square ODEF
= 616 – 392 = 224 cm2
Yes, I agree with Aarthi statement.

## AP 10th Class Maths Chapter 10 Important Questions Circles

These AP 10th Class Maths Chapter Wise Important Questions 10th Lesson Circles will help students prepare well for the exams.

## 10th Lesson Circles Class 10 Important Questions with Solutions

### 10th Class Maths Circles 1 Mark Important Questions

Question 1.
Define secant.
A line which intersects a circle in two distinct points is called a secant of the circle.

Question 2.
What is tangent ?
A line which touches the circle at only one point is called a tangent of the circle.

Question 3.
How many tangents can a circle have?
Infinitely many.

Question 4.
How many tangents can be draw to a circle from an external point ?
Two.

Question 5.
cm. The tangent at a point A on the circle acts the line through O to B such that AB = 15 cm. Find OB.
Solution:

In ΔAOB, ∠OAB = 90°
OA = 8 cm, AB = 15 cm, OB = ?
By Pythagoras theorem,
OB2 = OA2 + AB2 = 82 + 152
OB2 = 64 + 225 = 289 = 172
OB = 17 cm

Question 6.
In the given figure, AB = BC = 10 cm. If AC = 7 cm, then find the length of BP.

Solution:

Tangents drawn from an external point of a circle are equal
10 – x = y
x + y = 10 …… (1)
7 – y = 10 – x
x – y = 3 ……. (2)
(1) + (2) ⇒ 2x = 13
x = 6.5
∴ BP = 6.5 cm

Question 7.
In the given figure, AB is a tangent to the circle centered at O. If OA = 6 cm and ∠OAB = 30°, then find the radius of the circle.

Solution:

In Right ΔOBA, B = 90°
OB ⊥ AB
sin 30° = $$\frac{\mathrm{OB}}{\mathrm{OA}}$$
$$\frac{1}{2}$$ = $$\frac{O B}{15}$$
OB = 3 cm.

Question 8.
In the given figure, AC and AB are tangents to a circle centred at O. If ∠COD = 120°, then ∠BAO =

Solution:

In the figure ΔABO ≅ ΔACO
∠CAO = 180 – 90 – 60 = 30°
= 180° – 90° = 30°
∴ ∠CAO = ∠BAO = 30°

Question 9.
In figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ =

Solution:
In the figure POQT
∠P = ∠Q = 90°
∠POQ = 110°
∠PTQ = 360° – 110° – 90° – 90°
= 360° – 290° = 70°

Question 10.
In the given figure, O is the centre of the circle and PQ is the chord. If the tangent PR at P makes an angle of 50° with PQ, then find the measure of ∠POQ

Solution:
ΔPOQ is an Isosceles triangle
∠OPR = 90° as OP ⊥ PR
∠OPQ = 90° – 50° = 40°
∴ ∠OQP = 40 = ∠OPQ
∴ ∠POQ = 180° – 40° – 40° = 100°

Question 11.
Find the length of tangent drawn to a circle of radius 9 cm from a point 41 cm from the centtre.
Solution:

In the figure 412 = 92 + AB2
1681 – 81 = AB2 ⇒ AB2 = 1600
AB = 40 cm

Question 12.
A circle is of radius 3 cm. Find the distance between two of its parallel tangents.
Solution:

AO + BO = 3 + 3 = 6 cm

Question 13.
In Fig. ΔABC is circumscribing a circle, the length of BC is …………….. cm.

Solution:
The lengths of tangents drawn from an external point of a circle are equal.
PB = BQ = 3 cm
AR = PA = 4 cm
CR= 11 – 4 = 7 cm
∴ QC = 7 cm
BC = BQ + QC
= 3 + 7 = 10 cm

Question 14.
The length of the tangent from an external point A to a circle of radius 3 cm, is 4 cm. Find the distance of A from the center of the circle.
Solution:

OB ⊥ AB
OA2 = 32 + 42
OA2 = 25
OA = $$\sqrt{25}$$ = 5 cm

Question 15.
Assertion (A) : A tangent to a circle is perpendicular to the radius through the point of contact.
Reason (R) : The lengths of tangents drawn from an external point to a circle are equal.

A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true.
Solution:
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Question 16.
In the given circle in fig., find the number of tangents parallel to tangent PQ.

Solution:
1 tangent

Question 17.
In the given figure, PA and PB are tangents from external point P to a circle with centre C and Q is any (joint on the circle. Then find the measure of ∠AQB.

Solution:
∠AQB = 180° – 55° = 125°

Question 18.
In the figure, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB =

Solution:
∠AOB = 180° – 50° = 130°
Let ∠OAB = ∠OBA = x°
x° + x° + 130° = 180°
2x° = 180° – 130°
2x = 50°
x = $$\frac{50}{2}$$
x = 25°
∠OAB = 25°

Question 19.
In the given figure, find the perimeter of ΔABC.

Solution:
The length of tangents drawn from an external point of a circle an equal.
BR = BP = 6 cm
AQ = AR = 5 cm
PC = QC = 4 cm
Perimeter of ΔABC = 10 + 9 + 11 =30 cm

Question 20.
In the given figure, BC and BD are tangents to the circle with centre O and radius 9 cm. If OB = 15 cm, then find the length (BC + BD).

Solution:
In ΔBCO, ∠OCB = 90°
152 = 92 + BC
225 – 81 = BC2
144 = BC2
BC = 12 cm
The lengths of tangents drawn from an external point are equal
BC = BD = 12 cm
BC + BD = 12 + 12 = 24 cm.

Question 21.
Assertion (A) : A tangent to a circle is perpendicular to the radius through the point of contact.
Reason (R) : The lengths of tangents drawn from the external point to a circle are equal.

A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true.
A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Question 22.
Statement p: If a circle touches all four Sides of a quadrilateral ABCD, then AB + CD = BC + AD.
Statement q : A parallelogram circum-scribing a circle is a rhombus.

A) p true, q false
B) p false, q true
C) p true, q true
D) p false, q false
C) p true, q true

Question 23.
In the figure, there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5 cm then PS = ……………….. cm.

Solution:
The lengths of tangents drawn from the external point of a circle are equal
PQ = PR = 5 cm
OQ ⊥ PS
∴ PQ = QS
∴ PQ = QS = 5 cm
PS = PQ + QS
= 5 + 5 = 10 cm

Question 24.
In the figure PA and PB are tangents to a circle with centre ‘O’. If ∠AOB = 120° then ∠OPA = ……………..

Solution:
In the figure, ∠AOB = 120°
∠AOP = ∠BOP = $$\frac{120}{2}$$ = 60°
OA⊥PA
∠OAP = 90°
∴ ∠OPA = 180° – 60° – 90°
= 180° – 150° = 30°

Question 25.
Assertion (A) : In the given figure PA = PB
Reason (R) : Area of Circle = πr

A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true.
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Question 26.
Assertion (A) : In the figure AB = 14 cm

Reason (R) : The lengths of tangents drawn from an external point of a circle are equal.
A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true.
A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Question 27.
Assertion (A) : In the figure l|| m

Reason (R) : The tangents drawn at the
end point of diameter of a circle are parallel.
A) Both Assertion (A) and Reason (R) are true and Reason, (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true.
Solution:
A) Both Assertion (A) and Reason (R) are true and Reason, (R) is the correct explanation of Assertion (A).

Question 28.
Assertion (A) : In the figure, x = 50°.

Reason (R) : Cyclic parallelogram is triangle.
A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
C) Assertion (A) is true but Reason (R) is false.
D) Assertion (A) is false but Reason (R) is true.
Solution:
C) Assertion (A) is true but Reason (R) is false.

Question 29.
For a circle which is inscribed in a AABC having sides 8 cm, 10 cm and 12 cm. Then match the column.

B) BE
C) CF

i) 1
ii) 7
iii) 5
iv) 3
A – (ii), B – (iii), C – (iv), D – (i)

Question 30.
If two tangents PA and PB are drawn to a circle with centre ‘O’ from an external point ‘P’ (figure), then match the column.

A) ∠PAB
B) ∠OAP
C) ∠OAB
D) ∠AOB

i) 90°
ii) θ/2
iii) 90 – θ/2
iv) 180° – θ
A – (iii), B – (i), C – (ii), D – (iv)

Question 31.
If AB is a chord of length 6 cm, of a circle of radius 5 cm, the tangents at A and B intersects at a point X, then match the column.

A) AY
B) OY
C) XA
D) OA

i) 4 cm
ii) 3.75cm
iii) 5 cm
iv) 3 cm
A – (iv), B – (i), C – (ii), D

Question 32.
The length of the minutes hand of a clock is 7 cm then how much distance does it cover in one hour ?
44 cm

Question 33.
How many tangents can be drawn on a circle from a point outside the circle?
Solution:

Only two tangents can be drawn from an external point to the circle i.e., PA, PB are the two tangents to the circle.

Question 34.
What is the angle between the radius and tangent at the point of contact ?
90°

Question 35.
Statement p : If a circle touches all four sides of a quadrilateral ABCD, then AB + CD = BC + AD.
Statement q: A parallelogram circumscribing a circle is a rhombus.
A) p true, q false
B) p false, q true
C) p true, q true
D) p false, q false
C) p true, q true

Question 36.
The number of common tangents can be drawn to two concentric circles is ……………….
Solution:
0 (zero).

Question 37.
Statement ‘p’ : A tangent to a circle intersects it in one point.
Statement ‘q’ : We can draw infinite tangents to a given circle.

A) Statement ‘p’ and ‘q’ both are true.
B) Statement ‘p’ is true, ‘q’ is false.
C) Statement ‘p’ is false, ‘q’ is true.
D) Statement ‘p’ and ‘q’ both are false.
A) Statement ‘p’ and ‘q’ both are true.

### 10th Class Maths Circles 2 Mark Important Questions

Question 1.
If the tangent at a (mint P to a circle with centre O cuts a line through O at Q such that PQ = 24 cm and OQ = 25 cm. Find the radius of the circle.
Solution:
In ΔPOQ, ∠OPQ = 90°
OQ = 25 cm, PQ = 24 cm, OP = ?

By Pythagoras theorem,
OQ2 – OP2 + PQ2
OP2 = OQ2 – PQ2 = 252 – 242
OP2 = 72
∴ OP = 7 cm
Therefore, radius of the circle OP = 7 cm.

Question 2.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q. So that OQ = 13 cm. Find the length of PQ.
Solution:
In ΔOPQ, ∠OPQ = 90°
OP = 5 cm, OQ = 13 cm, PQ = ?

By Pythagoras theorem,
OP2 + PQ2 = OQ2
PQ2 – OQ2 – OP2
= 132 – 52 = 169 – 25 = 144
PQ2 – 122
∴ PQ = 12 cm

Question 3.
In the given figure, PQ is a chord of the circle centered at O. PT is a tangent to the circle at P. If ∠QPT = 55°, then find ∠PRQ.

Solution:
Given PQ is a chord.
‘O’ is centre of circle
PT is a tangent
∠QPT = 55°
∠QPT + ∠OPT = 90°
∠OPT = 90° – 55° – 35°
In ΔPOQ
∠POQ = 180° – ∠OPQ – ∠PQO
= 180° – 35° – 35° = 180° – 70° = 110°
Major curve ∠POQ = 360° -110° = 250°
∠QRP = $$\frac{1}{2}$$ × 250 = 125°
∴ ∠PRQ = 125°

Question 4.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:

In the figure
‘O’ is centre of circle
OA = 4 cm
OP = 5 cm
52 = 42 + PA2
25 – 16 = PA2
PA2 = 9
PA = √ 9 = 3 cm
∴ The length of tangent = 3 cm

Question 5.
In the given figure, tangents AB and AC are drawn to a circle centred at O. If ∠OAB = 60° and OB = 5 cm, find lengths OA and AC.

Solution:
In the figure,
AB and AC are tangents
AB = AC
∠OAB = 60°
OB = 5 cm
‘O’ is centre of circle.
OC = 5 cm

In Right ΔOBA, ∠B = 90°
sin 60° = $$\frac{5}{\mathrm{OA}}$$
$$\frac{\sqrt{3}}{2}$$ = $$\frac{5}{\mathrm{OA}}$$
√3 OA = 10
OA = $$\frac{10}{\sqrt{3}}$$ cm
tan 60° = $$\frac{5}{\mathrm{OA}}$$
√3 = $$\frac{5}{\mathrm{OA}}$$
AB = $$\frac{5}{\sqrt{3}}$$
AB = AC
∴ AC = $$\frac{5}{\sqrt{3}}$$ × $$\frac{\sqrt{3}}{\sqrt{3}}$$ = $$\frac{5}{3}$$√3 cm
∴ OA = $$\frac{10}{\sqrt{3}}$$ × $$\frac{\sqrt{3}}{\sqrt{3}}$$ = $$\frac{10 \sqrt{3}}{3}$$ cm
AC = $$\frac{5}{3}$$√3 cm

Question 6.
In fig. AB is diameter of a circle centred at O. BC is tangent to the circle at B. if OP bisects the chord AD and ∠AOP = 60°, then find ∠C.

Solution:

‘O’ is centre of circle
BC is tangent
∴ AP = PD
∠AOP = 60°
∠APO = 90°
∴ ∠BAC = 180° -60° -90°
= 180° – 150° = 30°
OB ⊥ BC
∠OBC = 90°
In ΔABC
∠A + ∠B + ∠C = 180°
30° + 90° + ∠C = 180°
120° + ∠C = 180°
∠C = 180° -120°
∠C = 60°
C = 60°

Question 7.
In fig. XAY is a tangent to the circle centred at O. If ∠ABO = 40°, then find ∠BAY and ∠AOB.

Solution:
Given XAY is a tangent to the circle
‘O’ is centre of circle.
∠ABO = 40°
OA = OB
∴ ∠OAB = ∠OBA = 40°
∠AOB = 180°- 40°- 40°
= 180° – 80° = 100°
∠AOB = 100°
OA⊥XY
∴ ∠OAY = 90°
∠OAB + ∠BAY = 90°
40° + ∠BAY = 90°
∠BAY = 90°-40°
∠BAY = 50°
∴ ∠BAY = 50°

Question 8.
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

Solution:
Given TP and TQ are two tangents
‘O’ is centre of circle
we have to prove ∠PTQ = 2∠OPQ

We have TP = TQ
∴ TPQ is an Isosceles Triangle
∠TPQ = ∠TQP
= $$\frac{1}{2}$$(180 – θ) = 90 – $$\frac{\theta}{2}$$
∠OPT = 90°
∠OPQ = ∠OPT – ∠TPQ
= 90° – (90° – $$\frac{\theta}{2}$$) = $$\frac{1}{2}$$θ = $$\frac{1}{2}$$∠PTQ
∴ ∠PTQ = 2∠OPQ.

Question 9.
The distance between two tangents parallel to each other of a circle is 13 cm. Find the radius of the circle.
Solution:

Distance between parallel tangents = 13 cm
∴ Diameter =13 cm
Radius = $$\frac{13}{2}$$ = 6.5 cm

Question 10.
In the given figure, PA is a tangent to the circle drawn from the external point P and PBC is the secant to the circle with BC as diameter. If ∠AOC = 130°, then find the measure of ∠APB, where O is the centre of the circle.

Solution:
In the figure ∠AOC = 130°
‘O’ is centre of circle
∠AOP = 180° – 130° = 50°
OA ⊥ PA
∠OAP = 90°
In ΔOAP,
∠OAP + ∠AOP + ∠APO = 180°
90° + 50° + ∠APO = 180°
∠APO = 180°-140°
∠APO = 40°
∴ ∠APB = 40°

Question 11.
In Fig. perimeter of ΔPQR p is 20 cm. Find the length of tangent PA.

Solution:
Given perimeter of ΔPQR = 20 cm
PQ + QR + PR = 20 cm
PQ + (QC + CR) + PR = 20
The lengths of tangents drawn from external point of circle are equal.

(PQ + QA) + RB + PR ) = 20 (by (1))
PA + PB = 20
PA + PA = 20 (by (1))
2PA = 20
PA = 10 cm

Question 12.
In Fig. BC is tangent to the circle at point B of circle centered at O. BD is a chord of the circle so that ∠BAD = 55°. Find m ∠DBC.

Solution:
Given BC is tangent to the circle.
Let point B of circle.
O’ is a centre of circle BD is a chord.
Angle in a semi circle is 90°
∠DBA = 180° – 90° – 55°
= 180 – 145 = 35°
∴ ∠DBC = 90° – 35° = 55°
∴ ∠DBC = 55°

Question 13.
XY and MN are the tangents drawn at the end points of the diameter DE of the circle with centre O. Prove that XY || MN.

Solution:
Given XY and MN are two tangents drawn at the end points of the diameter DE.
O is the centre of the circle.
We have to prove XY || MN .
We have
OE ⊥ MN
OD ⊥ XY
∴ ∠OEM = ∠OEN = 90°
∠ODX = ∠ODY = 90°
Also ∠ODY + ∠OEN = 90° + 90° = 180°
Sum of interior angles on the same side of the transversal is 180°.
∴ XY || MN.

Question 14.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other is a secant to the circle.
Solution:

Question 15.
PA is the tangent drawn to a circle of whose centre is ‘O’, OA is the radius and P is the external point of the circle. If PA = 24 cm, OP = 25 cm., then find its radius.
Solution:
Given PA = 24 cm, OP = 25 cm
We have OP2 = OA2 + PA2
⇒ (25)2 = OA2 + (24)2
⇒ 625 = OA2 + 576
⇒ OA2 = 49
⇒ OA = 7 cm

Question 16.
Find the length of the tangent from a point 13 cm away from the centre of the circle of radius 5 cm.
Solution:
Radius of the circle = 5 cm
Length of the tangent = x cm

Distance between centre to point be = 13 cm
132 = 52 + x2
x2 = 169 – 25 = 144 ⇒ x = 12 cm
Now, in ΔAOP,
∠POA + ∠OPA + ∠A = 180°
⇒ ∠POA + 40° + 90° = 180°
⇒ ∠POA = 50°

Question 17.
Find the length of the tangent from a point, which is 9.1 cm away frbm the centre of the circle, whose radius is 8.4 cm.
Solution:

Radius of the circle = r = 8.4 cm
Distance between centre to the point = d = 9.1 cm.
Length of the tangent = l = $$\sqrt{d^2-r^2}$$
= $$\sqrt{(9.1)^2-(8.4)^2}$$
= $$\sqrt{82.81-70.56}$$
= $$\sqrt{12.25}$$ = 3.5 cm.
∴ Length of the tangent = 3.5 cm.

Question 18.
“The length of the tangent from an external point ‘P’ to a circle with centre ‘O’ is always less than OP.” Is this statement true ? Give reasons.
Solution:

Δ OAP is right triangle.
OP is hypotenuse. AP is tangent to the circle at A.
∴ OP > AP (v Hypotenuse is longest side)
OP > length of the tangent
∴ Given statement is true.

Question 19.
The length of the tangent to a circle from a point 17 cm from its centre is 8 cm. Find the radius of the circle.
Solution:
Let PA is the length of the tangent PA = 8 cm

The distance of the external point from the centre OP = 17cm
Radius = OA = r cm.
∴ 172 = r2 + 82
⇒ r2 = 172 – 82 = 289 – 64 = 225
⇒ r = $$\sqrt{225}$$ = 15 cm

Question 20.
A point P is 25 cm from the centre O of the circle. The length of the tangent drawn from P to the circle is 24 cm. Find the radius of the circle.
Solution:
From right angled Δ AOP

OP2 = OA2 + AP2
(25)2 = OA2 + (24)2
625 = OA2 + 576
OA2 = 625 – 576 = 49 = 72
OA = 7 cm
∴ The radius of the circle is 7 cm.

Question 21.
Find the length of the tangent from a point 13 cm away from the centre of the circle of radius 5 cm.
Solution:
Radius of the circle = 5 cm
Length of the tangent = x cm

Distance between centre to point be
= 13 cm
132 = 52 + x2
x2 = 169 – 25 = 144 ⇒ x = 12 cm

Question 22.
In the given figure, ‘O’ is the centre of a circle, OQ is the radius and OQ = 5 cm. The length of the tangent drawn from external point to the circle PQ =12 cm, then find the distance between the points ‘O’ and ‘P’.

Solution:
The radius is perpendicular to the tangent at the point of contact i.e. ∠OQP = 90°

ΔOQP is a Right angled triangle
⇒ (OP)2 = (OQ)2 + (PQ)2
⇒ OP = $$\sqrt{(\mathrm{OQ})^2+(\mathrm{PQ})^2}$$ = $$\sqrt{(5)^2+(12)^2}$$
⇒ OP = $$\sqrt{25 + 144}$$ = $$\sqrt{169}$$ = 13 cm
∴ The distance between two points O and P = 13 cm

Question 23.
AOB is the diameter of a circle with centre ‘O’ and AC is a tangent to the circle at A. If ∠BOC = 130°, then find ∠ACO.

Solution:

Given that
AOB is diame’ter and centre is O
AC is a tangent
∠BOC =130° and point of contact at A.
We know that
The angle at point of contacat is 90°
The angle of straight line is 180°
∠AOC + ∠BOC = 180°
∠AOC + 130° = 180°
∠AOC = 180° – 130°
∠AOC = 50°
In a triangle
The sum of angles = 180°
∠AOC + ZOAC + ∠ACO = 180°
∠ACO + 90° + 50° = 180°
∠ACO = 180° – 140°
∠ACO = 40°
So, the angle ∠ACO is 40°

### 10th Class Maths Circles 4 Mark Important Questions

Question 1.
From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular bisector of AB.
Solution:
Given O is the centre of circle.
AP and PB are tangents drawn to a circle from P that is AP = BP and AB is the chord which meets OP at C.

In ΔAPC and ΔBPC,
PA = PB (Side)
∠APC = ∠BPC (Angle)
PC = PC (Side)
by S-A-S similarity ΔAPC ~ ΔBCP
By CPCT, ∠ACP = ∠BCP
∠ACP + ∠BCP = 180°
Therefore, ∠ACP = ∠BCP = $$\frac{180^{\circ}}{2}$$ = 90°
Hence, OP⊥AB
So, OP is the perpendicular bisector of AB.

Question 2.
O is the centre of the circle in the given figure. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral.
Solution:
In the figure OA⊥AB and OB⊥BP.

Radius is perpendicular to the tangent at its point of contact.
∠OAP = ∠OBP = 90°
So, ∠OAP +∠OBP = 180°
∠OAP + ∠APB + ∠OBP + ∠AOB = 360°
(∠OAP + ∠OBP) + ∠APB + ZAOB = 360°
180° + ∠APB + ∠AOB = 360°
∠APB + ∠AOB = 360° – 180° = 180°
So, the quadrilateral OAPB is cyclic.

Question 3.
If AB, AC, PQ are tangents to the circle and AB = 5 cm in the given figure. Find the perimeter of the triangle.
Solution:
Given AB, AC, PQ are tangents.

AB = 5 cm
Tangents drawn to a circle for the external point are equal.
That is AB = AC, BP = PX, QC = XQ
Perimeter of ΔAPQ
= AP + AQ + PQ
= AP + AQ + PX + XQ
= (AP + PX) + (AQ + XQ)
= AB + AC
But, AB = AC = 5 cm
Perimeter = AB + AC = 5 + 5 = 10 cm

Question 4.
In the given figure, ΔABC is a right triangle, right angle at B such that BC = 6 cm and AB = 8 cm. Find the radius of the in circle.
Solution:

Given in ΔABC, ∠B = 90°
AB = 8 cm, BC = 6 cm
Let radius of circle be x cm
So, OPBQ is a square
That is OP= OQ = PB = BQ = x cm
Tangents drawn to the circle from an external point are equal.
That is AP = AR, PPB = BQ = x, CQ = CR
AR = AP = AB – BP = 8 – x
CR = CQ = BC – BQ = 6 – x
AR + CR = AC = 8 – x + 6 – x = 14 – 2x
by Pythagoras theorem,
AC2 = AB2 + BC2
(14 – 2x)2 = 82 + 62 = 102
14 – 2x = 10 ⇒ -2x = 10 – 14
-2x = -4
x = $$\frac{4}{2}$$ = 2 cm
∴ Radius of the circle is 2 cm.

Question 5.
Two circles touch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively. Prove that TQ = TR.
Solution:
Given TP, TQ and TR are tangents to the two externally touching circles.

Tangents drawn to the circle from an external point are equal.
That is TP = TQ → (1)
TP = TR → (2)
From (1) and (2) TQ = TR.

Question 6.
In the given figure, AP and AQ are tangents from A to the circle with centre O. B is a point on the circle, then prove that AX + XB = AY + YB.

Solution:
Given in the figure, AP, AQ and XY are tangents drawn to the circle. Tangents drawn from external point are equal.
That is, AP = AQ
XP = XB
YQ = YB
Now, AP = AQ
AX + XP = AY + YQ
AX + XjB = AY + YB (∵ XP = XB, YQ = YB)

Question 7.
In the given figure, a circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

Solution:
Given circle touches all the four sides of a quadrilateral AB, BC, CD and DA at P, Q, R and S respectively.
We know that if a circle touches all the four sides of a quadrilateral ABCD then AB + CD = BC + AD.
Given AB = 6 cm, BC = 7 cm and CD = 4cm
6 + 4 = 7 + AD

Question 8.
In the given figure CP and CQ are tangents to a circle with centre O. AB is another tangent touching the circle at R. If CP = 11 cm and BC = 7 cm then find the length of BR.

Solution:
Given, CP, CQ and AB are tangents to the circle with centre O. CP = 11 cm, BC = 7 cm.
We know that, tangents drawn from external point C are equal.
That is, AP = AR, BQ = BR and CQ = CP
BC + BQ = CP
7 + BQ = 11
BQ = 11 – 7 = 4cm
BQ = BR = 4 cm
Therefore, BR = 4 cm

Question 9.
In the given figure ΔABC is circum-scribing a circle. Find the length of BC.

Solution:
Given ΔABC is circumscribing a circle.
AC = 11 cm, AR = 4 cm, BR = 3 cm
We know that tangents drawn from an external point are equal.
That is AR = AQ, BP = BR, CP = CQ
AC = AQ + CQ = 11 cm
AR + CQ = 11 cm (∵ AQ = AR)
4 + CQ = 11
CQ = 11 – 4 = 7 cm
But, CQ = CP = 7 cm
and BP = 3 cm
BC = BP + CP = 3 cm + 7 cm = 10 cm
Therefore, BC = 10 cm

Question 10.
Prove that the angle between the two tangents drawn from an external to circle is supplementary to the angle subtended by the line joining the points of contact at the centre.
Solution:

Let ‘O’ be the centre of circle
P is the external point
AB is the chord
OA⊥PA and OB⊥PB
∴ ∠OBD = 90°; ∠OAP = 90°
Sum of all interior angles = 360°
∠OAP + ∠OBP + ∠BOA + ∠APB = 360°
90° + 90° + ∠BOA + ∠APB – 360°
∠BOA + ∠APB = 180°
∴ The angle between the two tangents drawn from an external point to a circle supplementary to the angle subtended by the line segment.

Question 11.
ABC is an isosceles triangle with AB = AC, circumscribed about a circle. Prove that BC is bisected at E.

Solution:
We know that,
The lengths of tangents drawn from an external point of a circle are equal.

In Isosceles triangle ABC, AB = AC.
AB – AF = AC – AF
BD = CF
BE = CF [by (1)]
∴ BC is bisected at the point of contact.

Question 12.
In the given figure, the radii of two concentric circles are 13 cm and 8 cm. AB is diameter of the bigger circle. BD is the tangent to the smaller circle touching it at D. Find the length AD.

Solution:

Join OD and AE
∠ODB = 90° (OD⊥BE)
∠AEB = 90° (angle in semi circle)
OD || AE (corresponding angles)
AE = 2 × OD
AE = 2 × 8 = 16 cm
In Right ΔODB,
BD2 = 132 – 82 = 169 – 64 = 105
BD = $$\sqrt{105}$$ cm
DE = $$\sqrt{105}$$ cm

In Right ΔAED,
= 162 + ($$\sqrt{105}$$)2 = 256 + 105 = 361
AD = $$\sqrt{361}$$

Question 13.
P & Q are centres of circles of radii 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of the circle of radius x cm, which touches the above circle externally. Given that angle PRQ is 90°. Write an equation in x and solve it.
Solution:

In Right APQR, by pythagoras theorem
PQ2 = PR2 + QR2
172 = (x + 9)2 + (x + 2)2 [Aftering solving]
x2 + 11x – 102 = 0
x2 + 17x – 6x – 102 = 0
x(x + 17) – 6(x + 17) = 0
(x – 6) (x + 17) = 0
x = 6 or x = -17
x = 6 cm only
∴ x = 6 cm

Question 14.
If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, respectively, prove that
AQ = $$\frac{1}{2}$$(BC + CA + AB).
Solution:

Given : The circle touches the side BC of ΔABC at P.
The circle touches the extended sides
AB and AC of the triangles at Q and R.
To prove : AQ = $$\frac{1}{2}$$ (BC + CA + AB)
Proof : We know that, “the lengths of tangents drawn from an external point of a circle are equal”.
BP = BQ ….. (1)
CP = CR ….. (2)
AQ = AR …… (3)
Perimeter of ΔABC = AB + BC + AC
From the figure
BC = BP + PC
AB + BC + AC = AB + (BP + PC) + AC
= AB + BQ + CR + AC
From the figrue
AB + BQ = AQ
CR + AC = AR
BQ + CR = BC
AB + BC + AC = AQ + AR
∴ AB + BC + AC = AQ + AQ
2AQ = AB + BC + AC
∴ AQ = $$\frac{1}{2}$$[AB + BC + AC]

Question 15.
In the figure, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.

Solution:
In the figure join OT.
OT meets PQ at the point R
ΔTPQ is an isosceles .
TO is the angle bisector of ∠PTO
We have, the lengths of tangents drawn from external point of a circle are equal.
∴ TP = TQ
OT⊥PQ
OT bisects PQ
PR = RQ = 4 cm
OR = $$\sqrt{\mathrm{OP}^2-\mathrm{PR}^2}$$
OR = $$\sqrt{5^2-4^2}$$ = 3cm
∠TPR + ∠RPO = 90° (∵ ∠TPO = 90°)
∠TPR + ∠PTR = 90°
∠RPO = ∠PTR
ΔTRP ~ ΔRPO (AA rule)
$$\frac{\mathrm{TP}}{\mathrm{PO}}$$ = $$\frac{\mathrm{RP}}{\mathrm{RO}}$$
⇒ $$\frac{T P}{15}$$ = $$\frac{4}{3}$$
TP = $$\frac{20}{3}$$ cm

Question 16.
In given figure XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that
∠AOB = 90°

Solution:

O is centre of circle
Join OC
In ΔOPA and ΔOCA
AP = AC (Tangents from A)
AO = BO (Common side)
by SSS congruence rule
ΔOPA ≅ ∠OCA
by CPCT ∴ ∠POA = ∠COA …… (1)
similarly ∠QOB = ∠OCB
∠QOB = ∠COB …….. (2)
here POQ is a diameter of circle
∴ ∠POA + ∠COA +∠COB + ∠QOB = 180°
by (1) & (2)
2∠COA + 2∠COB = 180°
∠COA + ∠COB = 90°
∠AOB = 90°

Question 17.
The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D and intersecting the larger circle at P on producing. Find the length of AP.
Solution:
In the figure
‘O’ is centre of circle
OD⊥BD
∴ ∠ODB = 90° = ∠ODP
Let P be a point on bigger circle.
∴ ∠APB = 90°
In ΔABP and ΔOBD
∠APB = ∠ODB = 90°
∠ABP = ∠ODB (Common)
ΔABP ~ ΔOBD (AA similarly)

∴ $$\frac{\mathrm{AP}}{\mathrm{OD}}$$ = $$\frac{\mathrm{AB}}{\mathrm{OB}}$$
$$\frac{A P}{8}$$ = $$\frac{26}{13}$$
AP = 2 × 8
AP = 16 cm

Question 18.
If tangents PA and PB from a point P to a circle with centre ‘O’ are inclined to each other at angle of 80°, then find the measured of ∠POA.
Solution:

Between ΔAOP and ΔBOP
∠OAP = ∠OBP = 90° (∵ tangent and line from origin meet at 90° to each other)
OA = OB = radius of the circle = r (say) and OP is the common side
Hence, we can say ΔAOP = ΔBOP
Therefore, we can say
∠OPA = ∠OPB = $$\frac{1}{2}$$(∠APB)
= $$\frac{1}{2}$$(80°) = 40°
(∵ given that tangents PA and PB are inclined to each other by 80°)
Now, in ΔAOP,
∠POA + ∠OPA + ∠A = 180°
⇒ ∠POA + 40° + 90° = 180°
⇒ ∠POA = 50°

Question 19.
Prove that “in two concentric circles, a chord of the bigger circle, that touches the smaller circle is bisected at the point of contact with the smaller circle.”
Solution:
Proof : We are given two concentric circles C1 and C2 with centre O and a
chord AB of the larger circle C1, touching the smaller circle C, at the point P
(see figure) we need to prove that AP = PB. Join OP.

Then AB is a tangent to the circle C2 at P and OP is its radius.
Therefore, by Theorem 9.1
i. e., OP⊥AB
Now, ΔOAP and ΔOBP are congruent. This means AP = PB. Therefore, OP is . the bisector of the chord AB, as the perpendicular from the centre bisects the chord.

Question 20.
From an external point, two tangents are drawn to a circle. A line joining the external point and the center of the circle bisects the angle between the tangents. Is this true or not ? Justify your answer.
Solution:

Proof : Let PQ and PR be two tangents drawn from a point P outside of the circle with centre O. Join OQ and OR, triangle OQP and ORP are congruent because we know that,
∠OQP = ∠ORP – 90° (Theorem 9.1) i.e.,
OP⊥XY
OP is common.
This means ∠OPQ = ∠OPR (CPCT)
Therefore, OP is the angle bisector of ∠QPR.
Hence, the centre lies on the bisector of the angle between the two tangents.

Question 21.
AB is a chord of the circle and AOC is its diameter, such that ∠ACB = 60°. If AT is the tangent to the circle at the point A, then find the measure of ∠BAT.
Solution:
According to the data ∠ACB = 60°

AOC = diameter; AB = chord
AT is a tangent to the circle at A.
∴ ∠ACB = 90° (∵ semi-circle angle)
∠BAC + ∠ACB – 90°
∠BAC + 60° – 90°
⇒ ∠BAC = 30°
AT⊥AOC, ∠CAT = 90°
∠BAC + ∠BAT = 90°
30° + ∠BAT = 90°
⇒ ∠BAT = 60°

Question 22.
In the given figure, TA and TB are tangents to the circle with centre ‘O’. If ∠ATB = 80°, then find the measure of ∠ABT.

Solution:
In ΔTAB

TA = TB (v Length of the tangents drawn from the external point are equal)
⇒ ∠TBA = ∠TAB (∵ Angles opposite to the equal sides are equal)
⇒ Let ∠TBA = ∠TAB = x°
80° + x° + x° = 180° (∵ Sum of the angles in a triangle is 180°)
2x° = 180° – 80° = 100°
x° = 50°
∴ ∠ABT = 50°

Question 23.
In the given figure AB, AC and PQ are tangents to a circle and AB = 6 cm. Find the perimeter of ΔAPQ.

Solution:

We know that tangents drawn from an external point to a circle are equal in length.
∴ AB = AC
AP + PB = AQ + QC
AP + PX = AQ + QX (∵ PX = BP and QX = QC)
AP + PX = AQ + QX = 6 cm
Now, Perimeter of
ΔAPQ = AP + PQ + AQ = (AP + PX) + (QX + AQ)
= 64 + 6 = 12 cm
∴ Perimeter of ΔAPQ = 12 cm

Question 24.
A circle of radius 3 cm is inscribed in a triangle ABC and AF = 5 cm, BF = 3 cm as shown in the figure.

Somu said that the measure of the side AC is 17 cm. Do you agree ? Give reasons.
Solution:
AB = 8 cm, AD = 5 cm, BE = 3 cm
Let CD = CE = x
AC = 5 + x, BC = 3 + x

Since OE = OF = radius and
OF ⊥ AB, OE ⊥ BC
∴ AB ⊥ BC
From ΔABC
AC2 = AB2 + BC2
(5 + x)2 = (8)2 + (3 + x)2 (∵ a2 – b2 = (a + b) (a – b))
(8 + 2x)2 = 64
8 + 2x = 32
⇒ 2x = 24
⇒ x = 12
∴ AC = 5 + x = 5 + 12 = 17 cm

Question 25.
Draw the tangent to a given circle with centre O from a point R outside the circle. How many tangents can be drawn to the circle from that point ?
Solution:

Steps of construction :

1. Draw a circle with centre ‘O’,
2. Take a point P outside the circle. Join OP.
3. Draw the perpendicular bisector to OP which bisects it at M.
4. Taking M as centre and PM or MO as radius draw a circle. Let the circle intersects the given circle at A and B.
5. Join P to A and B.
6. PA and PB are the required tangents.
7. Only two tangents can be drawn.

(OR)

### 10th Class Maths Circles 8 Mark Important Questions

Question 1.
PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.
Solution:
Given PQ = 8 cm, OP = OQ = 5 cm
Let TP = TQ = a cm
and TR = b cm

OT is perpendicular bisector of PQ.
i.e., OR = QR = $$\frac{P Q}{15}$$ = $$\frac{\sqrt{8}}{2}$$ = 4 cm

In ΔOPR,
OP = OR2 + PR2
OR2 = OP2 – PR2 = 52 – 42
OR2 = 25 – 16 = 9 = 32
OR = 3 cm

In ΔTPR, TP2 = TR2 + PR2 → (1)
and in ΔOPT, OT2 = TP2 + OP2 → (2)
Put (1) in (2)
OT2 = (TR2 + PR2) + OP2
(b + 3)2 = b2 + 42 + 52

Question 2.
In the given figure if AB = AC prove that BE = EC.

Solution:
Given in ΔABC, AB = AC which circumscribed a circle.
Tangents drawn from an external point are equal.

(AD + BD) + CE = (AF + CF) + BE
AB + CE = AC + BE
M + CE = M + BE (∵ AB = AC)
CE = BE (or) BE – CE
So, point of contact E can bisect the base BC.

Question 3.
In the figure, the in circle of AABC touches the sides BC, CA and AB at D, E and F respectively. Show that AF + BD + CE – AE + BF + CD = $$\frac{1}{2}$$(Perimeter of ΔABC).

Solution:
Tangents drawn from external point are equal.

AF + BD + CE = AE + BF + CD
Perimeter of ΔABC
= AB + BC + AC
= (AF + FB) + (BD + DC) + (AE + CE)
= AF + AE + FB + BD + DC + CE
= AF + AF + BD 4 BD + CE + CE
= 2AF + 2BD + 2CE (∵ AE – AF, FB = BD CD = CE)
= 2(AF + BD + CE)
(OR)
2(AE + BF + CD)
$$\frac{1}{2}$$(Perimeter of ΔABC)
= AF + BD + CE = AE + BF + CD
Hence, AF + BD + CE = AE + BF + CD
= $$\frac{1}{2}$$(Perimeter of ΔABC)

Quetsion 4.
A circle is inscribed in a ΔABC having sides 8 cm, 10 cm and 12 cm as shown in the figure. Find AD, BE and CF.

Solution:
Given a circle is inscribed in a ΔABC.
AB = 8 cm, BC = 10 cm, AC = 12 cm
We know that tangents drawn from an external point to a circle are equal.
Let AD – AF = x cm
BD = BE = y cm and
CE = CF = z cm AB – AD + DB = 8 cm
x + y = 8 cm → (1)
BC = BE + CE = 10 cm
y + z = 10 cm → (2)
AC = AF + CF = 12 cm
x + z = 12 cm → (3)
by adding (1), (2) and (3)
(x + y) + (y + z) + (x + z) = 8 + 10 + 12
x + y + y + z + z + x = 30
2 (x + y + z) = 30
x + y + z = $$\frac{30}{2}$$ = 15 cm → (4)

Put z = 7 in (2) y + 7 = 10
y = 10 – 7 = 3 cm
Put y = 3 cm in (1) x + 3 = 8 cm
x = 8 – 3 = 5 cm
Therefore,
AD = x = 5 cm
BE = y = 3 cm
CF = z = 7 cm

Question 5.
Prove that the angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.
Solution:
Let AP and AQ are tangents drawn to a circle from an external point to a circle with centre O.

That is AP = AQ
In ΔOAP and ΔOAQ,
AP = AQ (Tangents from the external point)
OA = OA (Common side)
by side-side-side congruency,
ΔOAP ≅ ΔOAQ
by ∠PCT – ∠OAP = ∠OAQ and ∠AOP = ∠AOQ
∠OPA = ∠OQA = 90°
∠PAQ = 2, ∠OAP and ∠POQ = 2.∠AOP

In ΔAOP,
∠OAP + ∠AOP + ∠OPA = 180°
∠OAP + ∠AOP = 180° – 90° = 90°
∠AOP = 90° – ∠OAP
2.∠AOP = 2 (90 – ∠OAP)
∠POQ = 180° – 2.∠OAP (∵ 2∠AOP = ∠POQ)
∠POQ + 2.∠OAP = 180° (∵ 2∠OAP = ∠PAQ)
Hence proved, ∠POQ + ∠PAQ = 180°

Question 6.
In the figure, a circle is inscribed in a quadrilateral ABCD in which ∠B = 90°. If AD = 23 cm, AB = 29 cm and DS = 5 cm. Find the radius of the circle.

Solution:
Given the circle touches the quadrilateral ABCD at P, Q, R and S respectively, ∠B = 90° with centre O.
AB = 29 cm. AD = 23 cm and DS = 5 cm. We know that, tangents drawn from the external point are equal.
AP = AS, BP = BQ, CQ = CR and DR = DS
AD = AS + DS = 23 cm
AS = 23 – DS = 23 – 5 = 18 cm
But, AS = AP = 18 cm
AB = AP + PB = 29 cm
18 + PB = 29
PB = 29 – 18 = 11 cm
But OPBQ is a square.
Therefore, OP = OQ = BQ = PB = 11 cm
Hence, radius r = 11 cm

Question 7.
In the given figure there are two concentric circles with centre O of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles. If AP = 12 cm, find the length of BP.

Solution:
Given two concentric circles with centre O of radii. OA = 5 cm, OB = 3cm.
Tangent AP = 12 cm
In ΔOAP, ∠OAP = 90°
OP2 = OA2 + AP2
⇒ 52 + 122 = 169
OP = $$\sqrt{169}$$ = 13 cm

In ΔOBP, ∠OBP = 90°
OB2 + BP2 = OP2
32 + BP2 = 132
BP2 = 169 – 9 = 160
Therefore, BP = $$\sqrt{160}$$ = 4$$\sqrt{10}$$ cm

Quetsion 8.
Draw two tangents to a circle of radius 2.5 cm, from a point ‘P’ at a distance of 7 cm from its centre.
Solution:

1. Draw u line segment PO = 7 cm.
2. From the point O, draw a circle of radius = 2.5 cm.
3. Draw a perpendicular bisector of PO. Let M be the mid-point of PO.
4. Taking M as centre and 0M as radius, draw a circle.
5. Let this circle intersects the given circle at the point Q and R.
6. Join PQ and PR.

Question 9.
Two concentric circles of radii 10 cm and 6 cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.

Solution:
Radius of outer circle = R = 10 cm
Radius of inner circle = r = 6 cru
Given that AB in a chord of outer circle and tangent to inner circle of P.
Draw OP ⊥ AB
Then PA = PB
In Δ OAP,
∠OPA = 90°
OP2 + PA2 = OA2
62 + PA2 = 102
PA2 = 100 – 36 = 64
⇒ PA = 8 cm
∴ AB = PA + PB
= 2PA
= 2(8)
= 16 cm
Length of the chord AB = 16 cm

AP 10th Class Maths Important Questions Chapter 10 Tangents and Secants to a Circle

Question 1.
What do we call the part a and b in the below circle ?

Solution:
‘a’ is minor segment and ‘b’ is major segment.

Question 2.
Find the length of the tangent to a circle of radius 7 cm at a point from a distance 25 cm from the centre.
Solution:
Given OA = 25 cm, OB = r = 7 cm
In ΔAOB, ∠B = 90°

OA2 = OB2 + AB2
⇒ AB2 = OA2 – OB2 = 252 – 72
⇒AB = $$\sqrt{25^{2}-7^{2}}=\sqrt{625-49}$$
= $$\sqrt{576}$$
= 24 cm

Question 3.
Find the area of a sector of a circle whose radius is 7 cm and angle at the centre is 60°.
Solution:
Radius = 7 cm, Angle at centre = 60°
Area of the sector = $$\frac{\mathrm{x}}{360}$$ x πr²
= $$\frac{60 \times \frac{22}{7} \times 7 \times 7}{360}=\frac{154}{6}$$ = 25.66 cm2

Question 4.
A tangent is drawn to a circle of radius 4 cm. from a point that lies at a distance of 5 cm. from the centre. Find the measure of length of the tangent.
Solution:

ΔOAB is a right triangle.
OA2 = 0B2 + AB2
52 = OB2 + 42
OB2 = 25 -16 = 9
OB = $$\sqrt{9}$$ = 3 cm.

Question 5.
Find the area of the shaded part in the given figure.

Solution:
= Area of semi-circle – Area of triangle

Question 1.
Find the length of the tangent from a point 13 cm away from the centre of the circle of radius 5 cm.
Solution:
Radius of the circle = 5 cm
Length of the tangent = x cm

Distance between centre to point be = 13 cm
132 = 52 + x2
x2 = 169 – 25 = 144 ⇒ x = 12 cm

Question 2.
If tangents PA and PB from a point P to a circle with centre ‘O’ are inclined to each other at angle of 80°, then find the measured of ∠POA.
Solution:

Between ΔA0P and ΔB0P
∠OAP = ∠OBP = 90° (∵ tangent and line from origin meet at 90° to each other)
0A=0B = radius of the circle = r (say)
and OP is the common side
Hence, we can say ΔAOP = ΔBOP
Therefore, we can say
∠OPA = ∠OPB = $$\frac { 1 }{ 2 }$$(∠APB)
= $$\frac { 1 }{ 2 }$$(80°) = 40°
(∵ given that tangents PA and PB are inclined to each other by 80°)
Now, in ΔAOP,
∠POA + ∠OPA + ∠A = 180°
⇒ ∠POA + 40° + 90° = 180°
⇒ ∠POA = 50°

Question 3.
Draw a circle of radius 3 cm, mark a point ‘P’ on the circle and draw a tan gent at ‘P’.
Solution:
Steps of construction:

1) Draw a circle of radius ‘3’ cm from the centre “O” and pick a point ‘P’ on the circle. Join $$\overline{\mathrm{OP}}$$
2) Now draw a perpendicular at the point ‘P’ to the line segment $$\overline{\mathrm{OP}}$$
such that XY ⊥ $$\overline{\mathrm{OP}}$$.
3) Then $$\overline{\mathrm{XY}}$$ is the desired tangent at ‘P’ to the given circle.

Question 4.
A chord of a circle of radius 10 cm. subtends a right angle at the centre. Find the area of the corresponding minor segment (use π = 3.14),
Solution:
Radius of circle (r) = 10 cm
Sector angle (x) = 90°
Radius of sector (r) = 10 cm.

Area of sector OACB = $$\frac{\mathrm{x}}{360}$$ × πr²
= $$\frac{90^{\circ}}{360^{\circ}}$$ × 3.14 × 10 × 10
= 78.5 Sq. cm
Area of Δ AOB = $$\frac { 1 }{ 2 }$$ × 10 × 10
= 50 cm2
Area of the Minor Segment = Area of sector OACB – Area of Δ OAB
= 78.5 – 50.0 = 28.5 cm2

Question 1.
A chord of circle of radius 10 cm sub-tends a right angle at the centre. Find the area of the corresponding :
i) Minor segment ii) Major segment (use π = 3. 14)
Solution:
Radius of circle (r) = 10 cm
Sector angle (x) = 90°
Radius of sector (r) = 10 cm.

Area of sector OACB = $$\frac{\mathrm{x}}{360}$$ × πr2
= $$\frac{90^{\circ}}{360^{\circ}}$$ × 3.14 × 10 × 10
= 78.5 Sq. cm
Area of ∆ AOB = $$\frac { 1 }{ 2 }$$ × 10 × 10
= 50 cm2
Area of the Minor Segment = Area of sector OACB – Area of ∆ OAB
= 78.5 – 50.0 = 28.5 cm2
Area of Major Segment = Area of the circle – Area of the Minor Segment
= (3.14 × 10 × 10) – 28.5
= 314-28.5 = 285.5 cm2

Question 2.
Draw a circle of radius 3 cm. Take a point ’P’ at a distance of 5 cm from the centre of the circle. From P, draw 2 tangents to the circle.
Solution:
Award marks for construction as follows:

i) To draw a circle with radius 3 cm
ii) To plot a point P such that OP = 5 cm.
iii) To Bisect QP at M and draw circle with radius OM br MP
iv) To draw tangents from intersecting points of two circles.

Question 3.
Draw a Circle of radius 4 cm. From a point 7.5 cm away from its centre, construct the pair of tangents to the circle.
Solution:
1) Draw a circle with:radius 4 cm with centre O.
2) Locate a point P such that OP 7.5 cm
3) Bisect OP and draw circle with radius MO or MP with centre M.
4) Draw tangents PA and PB from external point P to the given circle.

Question 4.
Draw a circle of radius 5 cm. From a point 8 cm away from its centre, con-struct a pair of tangents to the circle. Find the lengths of tangents.
Solution:
Steps of construction :
1) Construct a circle with a radius of 5 cm.
2) Trace the point ‘p’ in the exterior of the circle which is at a distance of ‘8’ cm from its centre.
3) Construct a perpendicular bisector to OP which meets at M.
4) The draw a circle with a radius of MP or MO from the point M. This circle cuts the previous circle drawn from the centre ‘O’ at the points A and B.
5) Now join the points PA and then PB.
6) PA, PB are the required tangents which are measured 6.2 cm long.

OA = 5 cm ; OP = 8 cm
AP = PB = 6.2 cm.

Question 5.
Draw a circle of radius 4 cm and draw a pair of tangents to the circle, which are intersecting each other 6 cm away from the centre.
Solution:

Steps of Construction:
1) Draw a circle with centre ‘O’ and radius 4 cm.
Take a point ‘P’ outside the circle such that OP = 6 cm. Join OP.
Draw the perpendicular bisector to OP which bisects it at M.
4) Taking M as centre and PM or MO as radius draw a circle.
Let the circle intersects the given circle at ’A’ and ‘B’.
Join P to A and B.
PA and PB are the required tangents of lengths.

Question 6.
Two tangents TP and TQ are drawn to a circle with centre ‘O’ from an external point T, then prove that ∠PTQ = 2. ∠OPQ.
Solution:
Given a circle with centre 0.
Two tangents TP, TQ are drawn to the circle from an external point T.
We need to prove ∠PTQ = 2∠OPQ
Let ∠PTQ = θ

TP = TQ (The lengths of tangents drawn from an external point to a circle are equal)
So ΔTPQ is an isosceles triangle
∴ ∠TPQ + ∠TQP + ∠PTQ = 180°
(Sum of three angles in a triangle)
∠TPQ = ∠TQP = $$\frac { 1 }{ 2 }$$(180° – θ)
= 90°- $$\frac{\theta}{2}$$
∠OPQ = ∠OPT – ∠TPQ
= 90° – 0(90° – $$\frac{\theta}{2}$$) = $$\frac{\theta}{2}$$
= $$\frac { 1 }{ 2 }$$ ∠PTQ
∴∠PTQ = 2∠OPQ

Question 7.
Find the area of the segment shaded in the figure in which PQ = 12 cm, PR = 5 cm and QR is the diameter of the circle with centre ‘O’. (Take π = $$\frac{22}{7}$$)

Solution:
To find the area of the segment shaded in the given figure.
Here ‘PQ’ = 12 cm; ‘PR’ = 5 cm; ‘QR’ is diameter
Now PQOR is a semicircle then angle in a semicircle is 90°.
then ∠QPR = 90°
∴ ΔPQR is a right angled triangle
∴ Area of ΔPQR = $$\frac { 1 }{ 2 }$$ bh
= $$\frac { 1 }{ 2 }$$ × PQ × PR
= $$\frac { 1 }{ 2 }$$ × 12 × 5 =30 cm2 ……………(1)
Now the area of shaded part = area of semicircle – area of ΔPQR
= $$\frac { 1 }{ 2 }$$πrcm2 – 30 cm2 …………….(2)
In ΔPQR, QR2 = PQ2 + PR2
(from Pythagoras theorem)
QR2 = 122 + 52
= 144 + 25 = 169 = 132
∴ QR =13 then
Radius of the circle (r) = QO = $$\frac{\mathrm{Q} R}{2}$$
= $$\frac{13}{2}$$ = 6.5 cm
then area of semicircle
= $$\frac { 1 }{ 2 }$$πrcm2
= $$\frac{1}{2} \times \frac{22}{7} \times \frac{13}{2} \times \frac{13}{2}$$ = 66.39 cm2 …………….(3)
Now putting the values of (1) and (3) in (2) we get
Area of shaded part = (66.39 – 30)
= 36.39 cm2.

Question 8.
Draw two tangents to a circle of radius 2.5 cm, from a point ‘P’ at a distance of 7 cm from its centre.
Solution:
1. Draw a line segment PO = 7 cm.
2. From the point O, draw a circle of radius = 2.5 cm.
3. Draw a perpendicular bisector of PO. Let M be the mid-point of PO.
4. Taking M as centre and OM as radius, draw a circle.
5. Let this circle intersects the given circle at the point Q and R.
6. Join PQ and PR.

Question 1.
As shown in the figure, radius of the given circle is 21 cm and ∠AOB = 120°. Then find the area of segment AYB.

Solution:

Given radius of area = OA = OB = 21 cm
Now3 the angle at centre for the sector $$\widehat{\mathrm{OAB}}$$ = 120°.
Formula for area of sector = $$\frac{x}{360}$$ x πrcm2
= $$\frac{120}{360} \times \frac{22}{7}$$ × 21 × 21
= 22 × 21 = 462 cm2 ……………………… (1)

Now area of segment $$\widehat{\mathrm{AYB}}$$
= Area of sector – area of Δ OAB.
Let $$\overline{\mathrm{OD}}$$ is perpendicular to AB, then
∠AOB = $$\frac{120}{2}$$ = 60°
∴ sin 60° = $$\frac{\mathrm{AD}}{\mathrm{OA}}$$

∴ area of ΔOAB = $$\frac { 1 }{ 2 }$$ bh
= $$\frac { 1 }{ 2 }$$ × AB × OD = AD × OD
=  = 190.95 cm2 ………….. (2)
∴ area of segment $$\widehat{\mathrm{AYB}}$$
= 462 – 190.95
= 271.05 cm2
= 271.05 cm2

Question 2.
In a wall clock, length of minutes needle is 7 cm. Then find the area covered by it in 10 minutes of time.
Solution:
Length of minutes needle =(r) = 7 cm
We need to calculate the area covered by in 10 minutes of time = Area of sector.

Now the angle covered by it in 60 minutes = 360°
∴ In 10 minutes = $$\frac{360}{60}$$ × 10 = 60°
Area of sector =  × 7 × 7
= $$\frac{154}{6}=\frac{77}{3}$$
∴ Area covered by it in 10 minutes of time = $$\frac{77}{3}$$ cm2 = 25.66 cm2

Question 3.
Find the area of a right hexagon inscribed in a circle having 14 cm of radius.
Solution:
Radius of circle = OA = OB = OC = OD = OE = OF = 14 cm

∠AOB = $$\frac{360}{6}$$ = 60°
∴ In ΔAOB AO = BO = 14 cm
∠AOB = 60°
then ∠OAB = ∠OBA
(∵ Opposite to equal sides)
And ∠OAB + ∠OBA + 60 = 180
=> ∠OAB = ∠OBA = ∠AOB = 60°
Hence it is an equilateral triangle.
∴ OA = OB = AB = 14 cm
∴ Area of hexagon = 6 (Area of ΔAOB)
= 6. $$\frac{\sqrt{3}}{4}$$ a2
= 6. $$\frac{\sqrt{3}}{4}$$ × 14 × 14
Area of hexagon = 294$$\sqrt{3}$$ cm2

Question 4.
Four carrorp board pans are arranged as shown in figure. Radius of the pan is 3 cm each. Then find the area in between of them.
Solution:
Area in between 4 pans = Area of square ABCD formed by joining their centres – 4 (area of sector)

Now side of square ABCD = 3 + 3 = 6
Then area of square ABCD
= 6 × 6
= 36 cm2
and now area of sector = $$\frac{x}{360}$$ × πr2
Area of 4 sectors
= 4 × $$\frac{90}{360} \times \frac{22}{7}$$ × 3 × 3
= $$\frac{198}{7}$$ = 28.3 cm2
∴ Area in between 4 pans i.e., shaded
= 36 – 28.3 = 7.7 cm2

## AP 10th Class Maths Chapter 9 Important Questions Some Applications of Trigonometry

These AP 10th Class Maths Chapter Wise Important Questions 9th Lesson Some Applications of Trigonometry will help students prepare well for the exams.

## 9th Lesson Some Applications of Trigonometry Class 10 Important Questions with Solutions

### 10th Class Maths Some Applications of Trigonometry 1 Mark Important Questions

Question 1.
If the height of the tower is equal to the length of its shadow, then find the angle of elevation of the sun.
Solution:

tan θ = 1
tan θ = tan 45°
θ = 45°
∴ Angle of elevation of sun = 45°.

Question 2.
What is the angle of elevation of the top of a 15 ih high tower at a point 15√3 in away from the base of the tower?
Solution:

tanθ = $$\frac{15}{15 \sqrt{3}}$$ = $$\frac{1}{\sqrt{3}}$$
tanθ = tan 30°
θ = 30°
∴ The angle of elevation = 30°

Question 3.
In figure, the angle of elevation of the top of a tower AC from a point B on the ground is 60°. If the height of the tower is 20 m, find the distance of the point from the foot of the tower.

Solution:

tan 60° = $$\frac{20}{\mathrm{AB}}$$
√3 = $$\frac{20}{\mathrm{AB}}$$
AB = $$\frac{20}{\sqrt{3}}$$ m.
Distance between foot of the tower and point is $$\frac{20}{\sqrt{3}}$$ m.

Question 4.
The angle subtended by a vertical pole of height 100 m at a point on the ground 100√3 m from the base it has measure of

Solution:
tan P = $$\frac{100}{100 \sqrt{3}}$$
tan P = $$\frac{1}{\sqrt{3}}$$
tan P = tan 30°
P = 30°

Question 5.
The string of a kite in air is 50 m long and it makes an angle of 60° with the horizontal. Assuming the string to be straight, find the height of the kite from the ground.

Solution:
sin 60° = $$\frac{x}{50}$$
$$\frac{\sqrt{3}}{2}$$ = $$\frac{x}{50}$$
√3 = $$\frac{x}{25}$$
x = 25√3 m
The height of kite from the ground = 25√3 m

Question 6.
What is the angle of elevation of the top of a 30 m high tower at a point 30 m away from the base of the tower ?

Solution:
tan θ = $$\frac{30}{30}$$
tan θ = 1
tan θ = tan 45°
θ = 45°
∴ The angle of elevation = 45°

Question 7.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground where it makes an angle 30°. The distance between the foot of the tree to the point where the top touches the ground is 8m. Find the height of the tree from where it is broken.
Solution:

tan 30° = $$\frac{x}{8}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{x}{8}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{x}{8}$$
x = $$\frac{8}{\sqrt{3}}$$ $$\frac{\sqrt{3}}{\sqrt{3}}$$ = $$\frac{8 \sqrt{3}}{3}$$m
∴ Height of the tree from where it is broken = m

Question 8.
A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the
foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.
Solution:

tan 60° = $$\frac{x}{15}$$
√3 = $$\frac{x}{15}$$
x = 15√3 m.
∴ Height of tower = 15√3 m

Question 9.
A pole 6 m high casts a shadow 2√3 m long on the ground, then find the angle of elevation of the sun.
Solution:

tan θ = $$\frac{6}{2 \sqrt{3}}$$ = $$\frac{3}{\sqrt{3}}$$ = $$\frac{\sqrt{3} \cdot \sqrt{3}}{\sqrt{3}}$$
tan θ = √3
tan θ = tan 60°
θ = 60°
∴ Angle of elevation = 60

Question 10.
In the figure θ = _________

Solution:
tan θ = $$\frac{20}{20}$$
tan θ = 1
tan θ = tan 45°
θ = 45°

Question 11.
The ratio between the length of a pole and its shadow is 1 : 1 then find the angle of elevation of sun.

Solution:
tan θ = 1
tan θ = tan 45°
θ = 45°

Question 12.
The ratio between length of a tower and its shadow is √3 : 1 then find the angle of elevation of sun.
Solution:

tan θ = $$\frac{\sqrt{3}}{1}$$
tan θ = √3
tan θ = tan 60°
θ = 60°

Question 13.
A pole of height 1 m make a shadow on the horizontal line √3 m, then find the sun’s altitude.
A) 30°
B) 45°
C) 60°
D) 75°
Solution:
A) 30°

tan θ = $$\frac{1}{\sqrt{3}}$$
tan θ = tan 30°
θ = 30°

Question 14.
From the figure, θ = ……………..

Solution:
sin θ = $$\frac{2}{4}$$ = $$\frac{1}{2}$$
sin θ = sin 30°
θ = 30°

Question 15.
In the figure x = ………….. cm.

Solution:
x2 = 62 + 82
x2 = 100
x = 10 m

Question 16.
In the figure p = ………………. cm.

Solution:
x2 = 52 – 42 = 25 – 16 = 9
x = √9
x = 3 cm

Question 17.
In the figure AB || CD, then θ = ………………..

Solution:
60°

Question 18.
In the figure, MN || AB then α = ………………

Solution:
45°

Question 19.
In the figure x = ……………. cm

Solution:
tan 45° = $$\frac{x}{6}$$ ⇒ 1 = $$\frac{x}{6}$$ ⇒ x = 6cm

Question 20.
In the figure x = ……………. cm

Solution:
tan 45° = $$\frac{4.5}{x}$$ ⇒ 1 = $$\frac{4.5}{x}$$ ⇒ x = 4.5cm

Question 21.
In the figure, AB = …………….. cm.

Solution:
sin 30°= $$\frac{A B}{10}$$
$$\frac{1}{2}$$ = $$\frac{A B}{10}$$
AB = 5 cm

Question 22.
A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is …………… m.
Solution:

In ABC,
$$\frac{\mathrm{AB}}{\mathrm{AC}}$$ = cos 60°
$$\frac{A B}{15}$$ = $$\frac{1}{2}$$
AB = $$\frac{15}{2}$$m = 7.5m

Question 23.
Find the angle of elevation of the top of 10m a high tower at a point 10 m away from the base of the tower.
Solution:

tan θ = $$\frac{10}{10}$$
tan θ = 1
tan θ = tan 45°
θ = 45°
Required angle = 45°

Question 24.
In the figure AB || CD then x = …………….. cm.

Solution:
∠ABC = ∠BCD = 30°
sin30° = $$\frac{5}{x}$$
$$\frac{1}{2}$$ = $$\frac{5}{x}$$
x = 5 × 2
x = 10

Question 25.
If a pole 6m high throws shadow of 2√3 m then the angle of elevation of the sun is ………………. .
Solution:

tan θ = $$\frac{6}{2 \sqrt{3}}$$
tan θ = $$\frac{3}{\sqrt{3}}$$
tan θ = $$\frac{\sqrt{3} \cdot \sqrt{3}}{\sqrt{3}}$$
tan θ = √3
tan θ = tan 60°
θ = 60°
Angle of elevation = 60°

Question 26.
In the figure x = …………….. m.

Solution:
x = 10 + 9 = 19 m

Question 27.
In the figure y = ………………

Solution:
∠BDC = ∠BCD, y = 45°

Question 28.
In the figure x = ………………m.

Solution:
θ = 45° (A.I.A)
tan 45° = $$\frac{x}{10.2}$$
1 = $$\frac{x}{10.2}$$
x = 10.2 m

Question 29.
Assertion (A) : In the figure θ = 60°
Reason (R) : In ΔABC, if ∠B = 90° then ∠A + ∠C = 90°

A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) Assertion (A) is true, Reason (R) is true but R is not the correct explanation of A. A D
C) A is true, R is false.
D) A is false, R is true.
Solution:
D) A is false, R is true.

Question 30.
Assertion (A) : A pole is of height 6m makes same length of shadow on the ground then the sun’s angle of elevation is 45°.
Reason (R) : sin2θ + cos2θ = 1
A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) Assertion (A) is true, Reason (R) is true but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
B) Assertion (A) is true, Reason (R) is true but R is not the correct explanation of A.

Question 31.
Assertion (A) : In the figure x = 11 cm.
Reason (R) : cosec2θ – cot2θ = 1

A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) Assertion (A) is true, Reason (R) is true but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
B) Assertion (A) is true, Reason (R) is true but R is not the correct explanation of A.

Question 32.
Assertion (A) : In the figure θ = 60°
Reason (R) : sec2θ – tan2θ = 1

A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) Assertion (A) is true, Reason (R) is true but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
D) A is false, R is true.

Question 33.
Assertion (A) : The given figure represents sin example for angle of elevation.
Reason (R) : sinθ × cosecθ = -1.

A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) Assertion (A) is true, Reason (R) is true but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
C) A is true, R is false.

Question 34.
A tower of height 100 m casts a shadow of length 100√3 ni then what is the angle of elevation of the sun at that time ? (OR) In the given figure what is the value of angle θ?

Solution:
In Δ ABC
tan θ = $$\frac{\mathrm{BC}}{\mathrm{AB}}$$
tan θ = $$\frac{100}{100 \sqrt{3}}$$ = $$\frac{2}{\sqrt{3}}$$
θ = 30°

Question 35.
Name the ‘angle of depression’ from the figure given below in which ∠B = 90°.

Solution:
∠DAC

Question 36.
Sathwik is flying a kite with the support of string of length 100 m. which make an angle (If the string is tied) 60° with the ground, then height of the kite is ________
Solution:
50√3 m (or) 86.6 m.

Question 37.
At a particular time, if the angle of elevation of the sun is 45°, then the length of the shadow of a 5 m high tree is __________ .
A) 5√3 m
B) 10 m
C) 5 m
D) $$\frac{5}{\sqrt{3}}$$
Solution:
C) 5 m

Question 38.
Draw a rough diagram to the given situation.
“A person observed a top of a tree 10 m. away from its foot at the angle of elevation is 45°.”
Solution:

BC = Height of the tree.

Question 39.
Identify the angle of depression from the below figure.

Solution:
α

Question 40.
Draw a rough diagram to the following situation. “A person is flying a kite at an angle of elevation θ and the length of thread from his hand to kite is l.”
Solution:

Question 41.
“You are observing top of your school building at an angle of elevation 60° from a point which is at 20 metres distance from foot of the building”. Draw a rough diagram to the above situation.
Solution:

Question 42.
Draw the diagram for the following situations. Length of the shadow of a x meter high pole is y meter in the morning 7 o’ clock and the angle of elevation of the sunrays with the ground is ‘θ’.
Solution:

Question 43.
The given figure shows the observation of point ‘C’ from point A. Find the angle of depression from A.

Solution:
30°

Question 44.
Statement (A): If the length of shadow of a vertical pole is equal to its height, then the angle of elevation of the sun is 45°.
Statement (B) : According to pythagoras theorem, h2 = l2 + b2, where h = hypotenuse, l = Length and b = base.
i) Both A and B are true
ii) A is true, B is false
iii) A is false, B is true
iv) Both A and B are false
Solution:
i) Both A and B are true

Question 45.
From a window, ‘h’ m high above the ground, of a house in a street, the angles of elevation and depression of the top and bottom of another house on the opposite side of the street are α and β respectively, then match the column.

A) DB
B) DE
C) CE

i) h(1 +tan α cot β)
ii) h sin β
iii) h tan α cot β
iv) h cot β
A – (iv), B – (iii), C – (i), D (ii)

Question 46.
Jani observes Vanaja on the ground from the balcony of the first floor of a building at an angle of depression 60°. The height of the first floor of the building is 10 metres. Draw the diagram for this data.

AB = Height of the first floor

Question 47.
Draw the diagram for the following situations. Length of the shadow of a x meter high pole is y meter in the morning 7 o’ clock and the angle of elevation of the sunrays with the ground is ‘θ’.

### 10th Class Maths Some Applications of Trigonometry 2 Mark Important Questions

Question 1.
The angle of elevation of the top of a tower from a point on the ground which is 30 m away from the foot of the tower is 30°. Find the height of the tower.
Solution:
Let A be the point of observer, B be the foot of the tower and C be the top of the tower.

Angle of elevation ∠CAB = 30°
Let height of tower BC = h m
Distance between observer and foot of the tower AB = 30 m
In ΔABC, ∠B = 90°
tan ∠CAB = $$\frac{\text { Opposite side }}{\text { Adjacent side }}$$ = $$\frac{\mathrm{AB}}{\mathrm{AC}}$$
tan 30° = $$\frac{\mathrm{h}}{30}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{\mathrm{h}}{30}$$
h = $$\frac{30}{\sqrt{3}}$$ × $$\frac{\sqrt{3}}{\sqrt{3}}$$ = $$\frac{30 \sqrt{3}}{3}$$ = 10√3
Therefore, height of the tower BC = 10√3 m

Question 2.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the angle made by the rope with the ground level is 30°.
Solution:
Let AB be the height of pole = h m
AC be the length of rope = 20 m
Angle of elevation ∠ACB = 30°

In ΔABC, ∠B = 90°
sin ∠ACB = $$\frac{\text { Opposite side }}{\text { Hypotenuse }}$$ = $$\frac{\mathrm{AB}}{\mathrm{AC}}$$
sin 30° = $$\frac{\mathrm{h}}{20}$$ (∵ sin 30° = $$\frac{1}{2}$$)
$$\frac{1}{2}$$ = $$\frac{\mathrm{h}}{20}$$ ⇒ h = $$\frac{1}{2}$$ × 20 = 10 m
Therefore, height of pole AB = 10 m

Question 3.
A bridge across a river makes an angle of 45° with the river bank. If the length of the bridge across the river is 150 m. What is the width of the river ?
Solution:
Let PQ be the length of bridge = 150 m
QR is the width of the river = x m
Angle of elevation ∠QPR = 45°

In ΔPQR, ∠R = 90°
sin ∠QPR = $$\frac{\text { Opposite side }}{\text { Hypotenuse }}$$ = $$\frac{\mathrm{QR}}{\mathrm{PQ}}$$
sin 45° = $$\frac{x}{150}$$
$$\frac{1}{\sqrt{2}}$$ = $$\frac{x}{150}$$
x = $$\frac{150}{\sqrt{2}}$$ $$\frac{\sqrt{2}}{\sqrt{2}}$$ = $$\frac{150 \sqrt{2}}{2}$$ = 75√2 m
Therefore, width of the river QR = 7√2 m

Question 4.
The height of tower is 20 m. What is the length of its shadow when sun’s altitude is 45° ?
Solution:
Let length of the tower AB = 20 m
Let length of the shadow BC = x units
Angle of elevation ∠ACB = θ = 45°
In ΔABC, ∠B = 90°

tan ∠ACB = $$\frac{\text { Opposite side }}{\text { Adjacent }}$$ = $$\frac{\mathrm{AB}}{\mathrm{BC}}$$
tan 45° = $$\frac{20}{x}$$
1 = $$\frac{20}{x}$$ (∵ tan 45° = 1)
∴ x = 20 m
Therefore, length of shadow BC = 20 m.

Question 5.
If the ratio of the height of tower and the length of the shadow is √3 : 1. What is the angle of elevation of the sun ?
Solution:

Given ratio of the height of tower and its shadow = AB = BC = √3 : 1
i.e., $$\frac{\mathrm{AB}}{\mathrm{BC}}$$ = $$\frac{\sqrt{3}}{1}$$
In ΔABC,
∠B = 90°
tan θ = $$\frac{\text { Opposite side }}{\text { Adjacent }}$$ = $$\frac{\mathrm{AB}}{\mathrm{BC}}$$
tan θ = $$\frac{\sqrt{3}}{1}$$ = √3 = tan 60°
∴ θ = 60° (∵ $$\frac{\mathrm{AB}}{\mathrm{BC}}$$ = $$\frac{\sqrt{3}}{1}$$)
Therefore, angle of elevation = 60°

Question 6.
What is the angle of elevation of the sun when the length of the shadow of a vertical pole is equal to its height ?
Solution:
Given height of pole = Length of its shadow
Angle of elevation = θ
In ΔPQR, ∠Q = 90°

θ = 45° (∵ tan 45° = 1)
Therefore, angle of elevation = 45°.

Question 7.
Find the length of the shadow on the ground of a pole of height 18 m when angle of elevation θ of the sun is such that tan θ = $$\frac{6}{7}$$.
Solution:

Given tan θ = $$\frac{6}{7}$$
$$\frac{18}{x}$$ = $$\frac{6}{7}$$
6x = 18 × 7
x = $$\frac{18 \times 7}{6}$$
x = 3 × 7
x = 21 m
∴ The length of shadow = 21 m.

Question 8.
The angle of depression of a car parked on the road from the top of 150 m high tower is 30°. Find the distance of the car from the tower (in metres).
Solution:

Let AB is the height of the tower
Given angle of depression = 30°
In the figure
∠DBC = ∠ACB = 30° (A.I.A)
AC = x m
AB = 150 m 150
tan 30° = $$\frac{150}{x}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{150}{x}$$
x = 150√3 m
∴ Distance of the car from the tower = 150√3m.

Question 9.
From the top of a pole, the angle of elevation of the top of other pole is found to be equal to the angle of depression of the foot of the tower. First pole is of height 20 m. Find the distance between the two poles.
Solution:

In the figure, θ = 45°
tan θ = $$\frac{20}{x}$$
tan $$\frac{20}{x}$$
1 = $$\frac{20}{x}$$
x = 20 m
∴ Distance between two poles = 20m.

Question 10.
A tower of height 60 m, casts a shadows 20√3 m long on the ground, then what is the angle of elevation of the sun ?
Solution:

Let AB = 60 m
BC = 20√3 m.
tan θ = $$\frac{60}{20 \sqrt{3}}$$ = $$\frac{3}{\sqrt{3}}$$
tan = tan 60° ⇒ θ = 60°

Question 11.
A boy observed the top of an electric pole at an angle of elevation of 30°, when the observation point is 10 meters away from the foot of the pole. Draw suitable diagram for the above situation.
Solution:

AB = Height of the Electric Pole
AC = Distance between the foot of the pole to the observer = 10 m
Angle of Elevation = 30°

Question 12.
A flag pole 4 m tall casts a 6 m shadow. At the same time, a nearby building casts a shadow of 24. m. How tall is the building?
Solution:

$$\frac{4}{x}$$ = $$\frac{6}{24}$$
6x = 24 × 4 ⇒ x = $$\frac{24 \times 4}{6}$$ = 16 m.
∴ The height of the building = 16 cm.

Question 13.
Rehman observed the top of a temple at an angle of elevation of 30°, when the observation point is 24 m. away from the foot of the temple. Find the height of the temple.
Solution:

In ΔABC
tan 30° = $$\frac{\mathrm{AB}}{\mathrm{AC}}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{\mathrm{h}}{30}$$
⇒ h = $$\frac{24}{\sqrt{3}}$$ × $$\frac{\sqrt{3}}{\sqrt{3}}$$ = 8√3 m

Question 14.
A flag pole 7 m tall, casts a 8 m shadow. At the same time, a nearby building casts a shadow of 32 mts. How tall is the building ?
Solution:

ΔABC ~ ΔDEF
$$\frac{\mathrm{AB}}{\mathrm{DE}}$$ = $$\frac{\mathrm{BC}}{\mathrm{EF}}$$ ⇒  = $$\frac{8}{32}$$
∴ DE = 28 m

Question 15.
A person from the top of a building of height 25 m has observed another building top and bottom at an angle of elevation 45° and at an angle of depression 60° respectively. Draw a diagram for this data.
Solution:

Question 16.
A person observed the top of a tree at an angle of elevation of 60° when the observation point was 5 m away from the foot of the tree. Draw a diagram for this data.
Solution:

From the figure,
BC = height of the tree = h m.
‘A’ is the observation point.
AB = Distance between tree to the observation point.
Angle of elevation is 60°.

Question 17.
If the angle of elevation of sun increases from ‘O’ to 90 then the length of shadow of a tower decreases. Is this true ? Justify your answer.
Solution:
Yes, this statement is true.
We observe this in day to day life.

Question 18.
If a tower of height ’h’ is observed from a point with a distance’d’ and angle ‘θ’, then express the relation among h, d and θ.
Solution:
For writing relation between h, θ, d as tan θ = $$\frac{\mathrm{h}}{\mathrm{d}}$$

Question 19.
“The top of a tower is observed at an angle of elevation 45° and the foot of the tower is at a distance of 30 metres from the observer”. Draw a suitable diagram for this data.
Solution:

C – Point of observer
AB – Tower
A – Top of the tower

Question 20.
“An observer standing at a distance of 10m from the foot of a tower, observes its top with an angle of elevation of 60°.” Draw a suitable diagram for this situation.
Solution:

AB – Tower
C – Position of observer
60° – Angle of elevation

Question 21.
A boat has to cross a river. It crosses river by making an angle of 60° with bank, due to the stream of river it travels a distance of 450 m to reach another side of river. Draw a diagram to this data.
Solution:

AB – width of river
AC – The distance travelled in river = 450 m
A – initial point, C – terminal point

Question 22.
A person 25 mts away from a cell – tower observes the top of cell – tower at an angle of elevation 30°. Draw the suitable diagram for this situation.
Solution:
AB = cell tower
BC = distance between
observer and cell tower

Question 23.
A straight highway leads to foot of the tower. A man standing at the top of the tower observes a car at an angle of depression of θ, which is approaching the foot of the tower with a uniform speed. Six seconds later the angle of depression is Φ. Draw a diagram for this data and analyze.
Solution:

Question 24.
From the top of a tower of h m height, Anusha observes the angles of depression of two points X and Y on the same side of tower on the ground to be α and β. Draw the suitable figure for the given information.
Solution:

Question 25.
From the top of the building, the angle of elevation of the top of a T.V. tower is α° and the angle of depression to its (T.V. tower) foot is β°. If distance of the building from the tower is ‘d’ meters, draw the suitable diagram to the given data.

Question 26.
The angle of elevation of the top of a tower from a point on the ground, which is 50 m away from the foot of the tower is 45°. Draw the diagram for the situation.

### 10th Class Maths Some Applications of Trigonometry 4 Mark Important Questions

Question 1.
The horizontal distance between two towers is 140 m. The angle of elevation of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 60 m. Find the height of the first tower.
Solution:
Let AB be the height of first tower and
CD be the height of second tower.

Let BE = x m
AB = AE + BE = (60 + x) m
CD = AE = 60 m
Distance between two towers = AC = DE = 140 m
Angle of elevation at the top of the first tower ∠BDE = 30°
In ΔBDE, ∠BED = 90°
tan ∠BDE = $$\frac{\text { Opposite side }}{\text { Adjacent side }}$$ = $$\frac{\mathrm{AB}}{\mathrm{AC}}$$
tan 30° = $$\frac{x}{140}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{x}{140}$$
x = $$\frac{140}{\sqrt{3}}$$ m
Height of the first tower = AE + BE = 60 + x
Therefore, height of the first tower = (60 + $$\frac{140}{\sqrt{3}}$$)m

Question 2.
The length of a string between a kite and a point on the ground is 90 m. If the string makes θ with the ground level such that tan θ = $$\frac{15}{8}$$, how high is the kite ? Assume that there is no slack in the string.
Solution:
Let point of kite A and point of kite on the ground B, other end of the string C.
Length of string and point in the ground BC = 90 m

Let height of kite AB = h m
Angle of elevation ∠ACB = θ
and tan θ = $$\frac{15}{8}$$
In ΔABC, ∠B = 90°
tan ∠ACB = $$\frac{\text { Opposite side }}{\text { Adjacent side }}$$
tan θ = $$\frac{\mathrm{AB}}{\mathrm{BC}}$$
$$\frac{15}{8}$$ = $$\frac{\mathrm{h}}{90}$$
⇒ h = $$\frac{15}{8}$$ 90
= $$\frac{675}{4}$$ = 168.75 m
Therefore, height of the kite is 168.75 m.

Question 3.
From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°. If the bridge is at a height of 8 m from the banks, then the width of the river.

Solution:

Let AC is the height of the bridge.
AC = 8m
given angles of depression = 30C and 45°
AC ⊥ BD
In the figure.
∠EAB = ∠ABC = 45° = 0 (AIA)
tan θ = $$\frac{8}{x}$$
tan 45° = $$\frac{8}{x}$$
1 = $$\frac{8}{x}$$
x = 8 ……….. (1)
tan 30° = $$\frac{8}{y}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{8}{y}$$
y = 8√3 ………….. (2)
(1) + (2) ⇒ x + y = 8 + 8√3 = 8(1 + √3) m
∴ Width of the river = 8(1 + √3) m.

Question 4.
Two poles of heights 25 m and 35 m stand vertically on the ground. The tops of two poles are connected by wire, which is inclined to the horizontal at an angle of 30°. Find the length of the wire and the distance between the poles.
Solution:

In the figure
AB = height of first pole = 25 m
CD = height of second pole = 35 m
Let the length o| wire = x m
Distance between poles = y m
Let BE = AC = y m
CE = AB = 25 m.
DE = 35 – 25 = 10 m.
In Right triangle BDE
tan 30° = $$\frac{10}{y}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{10}{y}$$
y = 10√3 m
sin 30° = $$\frac{10}{x}$$
$$\frac{1}{2}$$ = $$\frac{10}{x}$$
x = 20m
∴ Length of the wire = 20 m
Distance between poles = 10√3 m .

Question 5.
As observed from the top of a light house 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 45°. Determine the distance travelled by the ship during this time. (Use √3 = 1.73)

Solution:

In the figure
AB = 100 m
BC = x m
CD = y m
∠EAD = ∠BDA = 30° (AIA)
∠EAC = ∠BCA = 45° (AIA)
In Right triangle, ABC
tan 45° = $$\frac{100}{x}$$
1 = $$\frac{100^{\circ}}{x}$$
x = 100m
In ΔABD, tan 30° = $$\frac{100}{x+y}$$
$$\frac{2}{\sqrt{3}}$$ = $$\frac{100}{x+y}$$
100√3 = 100 + y
100√3 – 100 = y
y = 100(√3 – 1)m
Distance travelled = 100 (√3 – 1)
= 100(1.732 – 1)
= 100 (0.732)
= 73.2 m

Question 6.
At a point of level ground, the angle of elevation of a vertical tower is, found to be a such that tan α = $$\frac{1}{3}$$. After walking 100 m towards the tower, the angle of elevation p becomes such that tan β = $$\frac{3}{4}$$. Find the height of the tower.
Solution:

In the figure CD = 100 m
AB y m = Height of tower
BD = x m
Let ∠ACD = α, ∠ADB = β.
tan α = $$\frac{y}{x+100}$$
$$\frac{1}{3}$$ = $$\frac{y}{x+100}$$
x + 100 = 3y
x – 3y = -100 ………… (1)
tan β = $$\frac{y}{x}$$
$$\frac{3}{4}$$ = $$\frac{y}{x}$$
3x = 4y
3x – 4y = 0 …………. (2)

y = 60 m
3x – 4(60) = 0
3x = 240
x = 80 m
∴ Height of the Tower = 80 m

Question 7.
In Figure, the angles of elevation of the top of a tower AB of height ‘h’m, from two points P and Q at a distance of x m and y m from the base of the tower respectively and in the same straight line with it, are 60° and 30°, respectively. Prove that h2 = xy.

Solution:
In right triangle BAP
tan 60° = $$\frac{h}{x}$$
√3 = $$\frac{h}{x}$$
h = √3x ………… (1)
In ΔABQ,
tan 30° = $$\frac{h}{y}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{h}{y}$$
h√3 = y
h = $$\frac{y}{\sqrt{3}}$$ …….. (2)
(1) × (3) ⇒ h . h = √3x.$$\frac{y}{\sqrt{3}}$$
h2 = xy

Question 8.
Two poles of equal heights are standing opposite each other on either side of the road of with 60 m. From a point P between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively, as shown in Figure, Find the height of the poles and distances of the point from the poles.

Solution:
In the figure
tan 60° = $$\frac{h}{x}$$
√3 = $$\frac{h}{x}$$
h = √3x …………. (1)
tan 30° = $$\frac{h}{y}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{h}{y}$$
h = $$\frac{y}{\sqrt{3}}$$ …………. (2)
(1) = (2)
$$\frac{y}{\sqrt{3}}$$ = √3x
y = 3x …………
In the figure AC = 60 m
x + y = 60
y = 60 – x ……………. (4)
∴ 60 – x = 3x (by (3))
60 = 3x + x
60 = 4x
x = $$\frac{60}{4}$$
x = 15 m
y = 60 – 15 = 45 m. (by (4))
y = 45 m
h = √3x (by (1))
h = √3 × 15 m
h = 15√3 m
∴ Height of poles = 15√3 m
Distance from poles = 15 m (or) 45 m

Question 9.
An observer of height 1.6 m is 11.2 m away from a palm tree. The angle of elevation of the top of tree from his eyes is 45°. What is the height of the palm tree ?
Solution:
Height of observer = 1.6 m = AB = DE
Distance from palm tree = 11.2 m = AE = BD

Angle of elevation = ∠CBD = 45°
In Δ BCD = tan 45° = $$\frac{\mathrm{CD}}{\mathrm{BD}}$$
⇒ 1 = $$\frac{C D}{11.2}$$ ⇒ CD = 11.2 m
∴ Height of palm tree = CE = CD + DE = 11.2 m + 1.6 m = 12.8 m.

Question 10.
Two poles of heights 6 m. and 11m. stand on a plane ground. If the distance between the feet of the poles is 12 m. find the distance between their tops.
Solution:
Given,
Height of first pole = AB = 6m.
Height of second pole = CD = 11 m

Distance between feet of poles = AC = 12 m
Distance between the tops of the pole, i.e., BD
So, BE = AC = 12 m
Similarly, AB = EC = 6m.
Now, DE = DC – EC = 11 – 6 = 5 m
BD2 = DE2 + BE2
= 52 + 122 – 25 + 144 = 169
∴ BD – 13 m

Question 11.
Suppose Siddu is shooting an arrow from the top of a building at an height of 6 m to a target on the ground at an angle of depression of 60°. What is the distance between Siddu and the object?
Solution:

Height of the building = 6 m
Angle of depression = Angle of elevation at the ground = 60°
Let the distance of the target from the shooting point = x m
Then from the figure
sin 60° = $$\frac{6}{x}$$ ⇒ $$\frac{\sqrt{3}}{2}$$ = $$\frac{6}{x}$$
⇒ x = $$\frac{6 \times 2}{\sqrt{3}}$$ = $$\frac{2 \times \sqrt{3} \times \sqrt{3} \times 2}{\sqrt{3}}$$ = 4√3
∴ Distance = 4√3 m or 4 × 1.732 = 6.928 m.

Question 12.
An observer flying in an aeroplane at an altitude of 900 m observes two ships in front of him, which are in the same direction at an angles of depression of 60° and 30° respectively. Find the distance between the two ships.
Solution:

In ΔABC
tan 60 = $$\frac{900}{x}$$
√3 = $$\frac{900}{x}$$ ⇒ x = $$\frac{900}{\sqrt{3}}$$ = 300√3
In ΔABD
tan 30 = $$\frac{900}{x + 100}$$ ⇒ $$\frac{1}{\sqrt{3}}$$ = $$\frac{900}{300 \sqrt{3}+\mathrm{d}}$$
d = 600√3 m

Question 13.
From the top of the building the angle of elevation of the top of the cell tower is 60° and the angle of depression to its foot is 45°, if the distance of the building from the tower is 30 meters, draw the suitable diagram to the given data.
Solution:
AB – Height of the building
CD – Height of the tower C

∠CAE – angle of eleva-tion = 60°
∠EAD – angle ol depres-sion = 45°
Distance between building and tower = BD = 30m

Question 14.
If two persons standing on either side of a tower of height 100 metres observes the top of it with angles of elevation of 60° and 45° respectively, then find the distance between the two persons. [Note : Consider the two persons and the tower are onvthe same line.]
Solution:
Tower height = 100 m
Angles of elevation = 60° and 45°

In ΔABP, tan 60° = $$\frac{\mathrm{AB}}{\mathrm{PB}}$$
$$\frac{\sqrt{3}}{1}$$ = $$\frac{100}{x}$$ ⇒ √3x = 100 ⇒ x = $$\frac{100}{\sqrt{3}}$$ m
In ΔABQ tan 45° = $$\frac{\mathrm{AB}}{\mathrm{BQ}}$$
$$\frac{1}{1}$$ = $$\frac{100}{y}$$ ⇒ y = 100 m
The distance between Two persons = x + y
= $$\frac{100}{\sqrt{3}}$$ + 100
= $$\frac{100(\sqrt{3}+1)}{\sqrt{3}}$$ m

Question 15.
An observer flying in an aeroplane at an altitude of 900 m observes two ships in front of him, which are in the same direction at an angles of depression of 60° and 30° respectively. Find the distance between the two ships.
Solution:

In ΔABC, tan 60 = $$\frac{900}{x}$$
√3 = $$\frac{900}{x}$$ ⇒ x = $$\frac{900}{\sqrt{3}}$$ = 300√3
In ΔABD, tan 30 = $$\frac{900}{x + d}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{900}{300 \sqrt{3}+d}$$
d = 600√3 m

Question 16.
Two poles are standing opposite to each other on the either side of the road which is 90 feet wide. The angle of elevation from bottom of first pole to top of second pole is 45°, the angle of elevation from bottom of second pole to top of first pole is 30°. Find the . heights of poles, (use √3 = 1.732).
Solution:
Let AB = x and CD = y
In Δ ABD, tan 30° = $$\frac{x}{90}$$
x = 90 × tan 30°
⇒ x = 90 × $$\frac{1}{\sqrt{3}}$$ = $$\frac{90 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$$ = 30√3

= 30(1.732) = 51.960 = 51.96 ft.
In Δ BDC, tan 45° = $$\frac{y}{90}$$
⇒ y = 90 × tan 45°
⇒ y = 90 × 1 ⇒ y = 90 ft
∴ Height of the first pole = 51.96 ft
Height of the second pole = 90ft

Question 17.
The angle of elevation of the top of a tower from two points at a distance of 4m and 9 m from the base of the tower and in the same straight line with it, are complementary. Prove that the height of the tower is 6 m.
Solution:

tan θ = $$\frac{\mathrm{h}}{9}$$ _______ (1)
tan (90 – θ) = $$\frac{\mathrm{h}}{4}$$ ⇒ cot θ = $$\frac{\mathrm{h}}{4}$$ ⇒ tan θ = $$\frac{4}{\mathrm{~h}}$$ _______ (2)
from () and (2) ⇒ $$\frac{\mathrm{h}}{9}$$ = $$\frac{4}{\mathrm{~h}}$$ ⇒ h2 = 36 ⇒ h = 6
∴ Height of the tower = 6m

### 10th Class Maths Some Applications of Trigonometry 8 Mark Important Questions

Question 1.
An electrician has to repair an electric fault on a pole of height 4 m. He needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use which when inclined at an angle of 60° to the horizontal would enable him to reach the required position ?
Solution:

Let P and R are the top and foot of the pole.
Let Q be the point 1.3 m below the top P of the pole PR.
Then QR = PR – PQ = 4 – 1.3 = 2.7 m
Length of the ladder = QS = d m
Angle of elevation at Q = ∠QSR = 60°
In ΔQRS, ∠R = 90°
Sin ∠QSR = $$\frac{\text { Opposite side }}{\text { Hypotenuse }}$$
Sin 60° = $$\frac{\mathrm{QR}}{\mathrm{QS}}$$
$$\frac{\sqrt{3}}{2}$$ = $$\frac{2.7}{\mathrm{~d}}$$
√3 d = 2.7 × 2
d = $$\frac{2.7 \times 2}{\sqrt{3}}$$ = $$\frac{5.4 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$$ = $$\frac{5.4 \times \sqrt{3}}{3}$$
= (1.8)√3 = $$\frac{18}{10}$$ √3
∴ Length of ladder QS = d = $$\frac{9}{5}$$ √3 cm

Question 2.
From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The angle of elevation of the top of a water tank is 45°. Find 0 height of the tower ii) the depth of the tank.
Solution:
i) Let A be the point on the ground 40 m away from the foot of the tower B.
Let C be the top of the tower.
Let BC = h m.
CD be the height of the tank = x m. Angle of the elevation at C = ∠CAB = 30° Angle of the elevation at D = ∠DAB = 45°
In Δ ABC, ∠B = 90°

tan ∠CAB = $$\frac{\text { Opposite side }}{\text { Adjacent }}$$ = $$\frac{\mathrm{BC}}{\mathrm{AB}}$$
tan 30° = $$\frac{\mathrm{h}}{40}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{\mathrm{h}}{40}$$ (∵ tan 30° = $$\frac{1}{\sqrt{3}}$$)
h = $$\frac{40}{\sqrt{3}}$$
Height of the tower
BC = $$\frac{40}{\sqrt{3}}$$ × $$\frac{\sqrt{3}}{\sqrt{3}}$$ = $$\frac{40}{3}$$ × √3m

ii) In ΔABD, ∠B = 90°, ∠DAB = 45°
tan ∠DAB = $$\frac{\mathrm{BD}}{\mathrm{AB}}$$ = $$\frac{B C+C D}{A B}$$
tan 45° = $$\frac{h+x}{40}$$
1 = $$\frac{h+x}{40}$$ (∵ tan 45° = 1)
h + x = 40
x = 40 – h
= 40 – $$\frac{40 \sqrt{3}}{3}$$
= 40(1 – $$\frac{\sqrt{3}}{3}$$)
∴ Depth of the tank = CD = x = 40($$\frac{3-\sqrt{3}}{3}$$)m.

Question 3.
The angles of elevation of the top of a tower from two points at distances a and b metres from the base and in the same straight line with it are complementary. Prove that the height of the tower is $$\sqrt{\mathbf{a b}}$$ metres.
Solution:

Let P and Q are two points on the ground.
R be the foot of the tower.
S be the top of the tower.
From P angle of elevation at the top of towers ∠SPR = θ.
From Q angle of the elevation at the top of tower ∠SQR = 90 – θ
PR = a metres
QR = b metres
In ΔPRS, ∠SPR = θ.
tan∠SPR = $$\frac{\mathrm{SR}}{\mathrm{PR}}$$
tan θ = $$\frac{\mathrm{h}}{\mathrm{a}}$$ → (1)
In ΔQRS, ∠SQR = 90 – θ
tan ∠SQR = $$\frac{\mathrm{SR}}{\mathrm{QR}}$$
tan(90 – θ) = $$\frac{\mathrm{SR}}{\mathrm{QR}}$$ (∵ tan (90 – θ) = cot θ
cot θ = $$\frac{\mathrm{h}}{\mathrm{b}}$$ → (2)
From (1) and (2)
tan θ.cot θ = $$\frac{\mathrm{h}}{\mathrm{a}}$$.$$\frac{\mathrm{h}}{\mathrm{b}}$$
1 = $$\frac{\mathrm{h}^2}{\mathrm{ab}}$$ (∵ tan θ.cot θ = 1)
h2 = ab
∴ Height of the tower = h = $$\sqrt{\mathbf{a b}}$$ m

Question 4.
A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards it. It is takes 12 minutes for the angle of depression to change from 30° to 45°. How soon after this, will the car reach the tower ?
Solution:
Let A and B are the two positions of car. C and D are the foot of the tower and top of the tower.
Time taken to move d distance = 12 minutes
Let height of the tower CD = h m
Let AB = d m, BC = x m

Angle of elevation at the top of tower from A = ∠DAC = 30°
Angle of elevation at the top of tower from B = ∠DBC = 45°
In ΔDAC, ∠DAC = 30°
tan∠DAC = $$\frac{\text { Opposite side }}{\text { Adjacent side }}$$ = $$\frac{\mathrm{CD}}{\mathrm{AC}}$$
tan 30° = $$\frac{C D}{A B+B C}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{h}{d+x}$$ (∵ tan 30° = $$\frac{1}{\sqrt{3}}$$)
h = $$\frac{\mathrm{d}+\mathrm{x}}{\sqrt{3}}$$ → (1)
In ΔDBC, ∠DBC = 45°
tan ∠DBC = $$\frac{\mathrm{CD}}{\mathrm{AC}}$$
tan 45° = $$\frac{h}{x}$$ (∵ tan 45° = 1)
1 = $$\frac{h}{x}$$
h = x → (2)
From (1) and (2)
$$\frac{d+x}{\sqrt{3}}$$ = x
d + x = x√3
d = x√3 – x
d = x (√3 – 1) m
Time taken to travel d = x (√3 – 1) metres is 12 minutes.
Time taken to travel x metres

∴ Time taken to reach the tower = 6(√3 + 1) minutes

Question 5.
Two pillars of equal height and on either side of a road, which is 100 m wide. The angles of elevation of the top of the pillars are 60° and 30° at a point on the road between the pillars. Find the position of the point between the pillars and the height of each pillar.
Solution:
Let A and B are the top and foot of the first pillar.

C and D are the foot and top of the second pillar.
Let height of the pillars AB = CD = h m. Let point of observer is P.
Let distance from the foot of the first pillar to the observer = BP = x m.
Distance between two pillars BC = 100 m.
Then distance from the foot of the second pillar to the observer = CP = BC – BP = (100 – x) m
Angle of elevation from P at A is ∠APB = 60°
Angie of elevation from P at B is ∠DPC = 30°

In ΔABP, ∠B = 90°
tan∠APB = $$\frac{\text { Opposite side }}{\text { Adjacent side }}$$ = $$\frac{\mathrm{AB}}{\mathrm{BP}}$$
tan 60° = $$\frac{h}{x}$$ (∵ tan 60° = √3)
√3 = $$\frac{h}{x}$$
h = x.√3 m → (1)

In ΔDCP, C = 90°
tan∠DPC = $$\frac{\mathrm{CD}}{\mathrm{PC}}$$
tan 30° = $$\frac{h}{100-x}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{h}{100-x}$$ (∵ tan 30° = $$\frac{1}{\sqrt{3}}$$)
h = $$\frac{100-x}{\sqrt{3}}$$ → (2)
From (1) and (2)
x√3 = $$\frac{100-x}{\sqrt{3}}$$
x√3.√3 = 100 – x
3x + x = 100
4x = 100
x = $$\frac{100}{4}$$ = 25 m 4
BP = 25 m,
PC – 100 – x = 100 – 25 = 75 m
So, observer is 25 m far from the first pillar.

Question 6.
An aeroplane when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of the two planes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the aeroplane at that instant.
Solution:
Let the positions of aeroplanes are A and B.
Point of plane on the ground C.
Let point of observer is P.

Height of the first plane AC = 4000 m
Distance between the two planes = AB = h m
then height of the second plane = BC = AC – AB = (4000 – h) m
Let distance between observer and plane on the ground = PC = d m.
Angle of elevation at first plane = ∠APC = 60°
Angle of elevation at second plane = ∠BPC – 45°

In ΔAPC, ∠C = 90°
tan∠APC = $$\frac{\text { Opposite side }}{\text { Adjacent side }}$$ = $$\frac{\mathrm{AC}}{\mathrm{PC}}$$
tan 60° = $$\frac{4000}{d}$$
√3 = $$\frac{4000}{d}$$ ⇒ √3d = 4000
d = $$\frac{4000}{\sqrt{3}}$$ → (1)

In ΔBPC, ∠C = 90°
tan∠BPC = $$\frac{4000-\mathrm{h}}{\mathrm{d}}$$
tan 45° = $$\frac{4000-\mathrm{h}}{\mathrm{d}}$$
1 = $$\frac{4000-\mathrm{h}}{\mathrm{d}}$$ ⇒ d = 4000 – h → (2)
From (1) and (2)
$$\frac{4000}{\sqrt{3}}$$ = 4000 – h
h + $$\frac{4000}{\sqrt{3}}$$ = 4000
h = 4000 – $$\frac{4000}{\sqrt{3}}$$
= 4000(1 – $$\frac{1}{\sqrt{3}}$$)
Distance between two planes = h = 4000($$\frac{\sqrt{3}-1}{\sqrt{3}}$$) m

Question 7.
The horizontal distance between two poles is 15 m. The angle of depression of the top of the first pole as seen from the top of the second pole Is 30°. If the height of the second pole is 24 m. Find the height of the first pole.
Solution:
Let A and B are top and foots of first pole.
C and D are the foot and tops of second pole.

Let height of first pole AB = CE = h m
Height of second pole CD = 24 m
Then DE = CD – CE = (24 – h) m.
Distance between two poles = BC = AE = 15 m
Angle of depression at the top of the first pole ∠XDA = ∠DAE = 30°
In ΔDAE, ∠E = 90°
tan∠DAE = $$\frac{\text { Opposite side }}{\text { Adjacent side }}$$ = $$\frac{\mathrm{DE}}{\mathrm{AE}}$$

= 24 – 5(1.732) = 24 – 8.66
∴ Height of first pole = 15.34 m.

Question 8.
An aeroplane at an altitude of 1200 m find that two ships are sailing towards it in the same direction. The angles of depression of the ships as observed from the aeroplane are 60° and 30° respectively. Find the distance between the two ships.
Solution:

Let A and B are the places of two ships and C be the place aeroplane on the ground and D be the place of plane.
Height of aeroplane CD = 1200 m.
Let distance between two ships A and B is AB = d1 m.
Let BC = d2 m.
Angle of depression at ship A is ∠DAC = ∠XDA = 30°
Angle of depression at ship B is ∠DBC = ∠XDB = 60°

In ΔDBC, ∠C = 90°
tan∠DBC = $$\frac{\text { Opposite side }}{\text { Adjacent side }}$$ = $$\frac{\mathrm{CD}}{\mathrm{BC}}$$
tan 60° = $$\frac{120}{\mathrm{~d}_2}$$ (∵ tan 60° = √3)
√3 = $$\frac{120}{\mathrm{~d}_2}$$
d2 = $$\frac{1200}{\sqrt{3}}$$ = $$\frac{1200 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$$ = $$\frac{1200 \sqrt{3}}{3}$$
d2 = 400√3 m → (1)

In ΔACD, ∠C = 90°
tan∠DAC = $$\frac{\mathrm{CD}}{\mathrm{AC}}$$
tan 30° = $$\frac{1200}{\mathrm{~d}_1+\mathrm{d}_2}$$ (∵ tan 30° = $$\frac{1}{\sqrt{3}}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{1200}{\mathrm{~d}_1+\mathrm{d}_2}$$
d1 + d2 = 1200√3
d1 = 1200 √3 – d2 (∵ d2 = 400√3)
= 1200 √3 – 400 √3
d1 = 800 √3
Therefore, distance between the two ships = 800 √3 m (or) 800 × 1.732 = 1385.6 m

Question 9.
A man standing on the dock of a ship, which is 10 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of hill as 30°. Calculate the distance of the hill from the ship and the height of the hill.
Solution:
Let foot of the ship A point of man on the ship is B. Foot of the hill C and top of the hill D.
Height of the ship AB = CE = 10 m

Let CD = h m
Height of hill CD = CE + DE = (10 + h) m
Let distance between hill from the ship = AC = BE = d m
Angle of elevation at the top of hill ∠EBD = 60°
Angle of depression at the foot of the hill.
∠ACB = ∠CBE = 30°

In ΔACB, ∠A = 90°
tan∠ACB = $$\frac{\text { Opposite side }}{\text { Adjacent side }}$$ = $$\frac{\mathrm{AB}}{\mathrm{AC}}$$
tan 30° = $$\frac{10}{\mathrm{~d}}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{10}{\mathrm{~d}}$$ (∵ tan 30° = $$\frac{1}{\sqrt{3}}$$)
d = 10√3 m

In ΔBDE, ∠E = 90°
tan∠DBE = $$\frac{\mathrm{DE}}{\mathrm{BE}}$$
tan 60° = $$\frac{h}{d}$$ (∵ tan 60° = √3)
√3 = $$\frac{h}{d}$$
h = √3 d = √3(10√3) (∵ d = 10√3)
h = √3 × √3 × 10 = 3 × 10 = 30 m
Therefore, height of the hill = CD = h + CE = 30 + 10 = 40 m
Distance between hill and the ship = 10√3 m (or) 10 (1.732) = 17.32 m.

Question 10.
As observed from the top of a 150 m tall lighthouse the angles of depression of two ships approaching it are 30° and 45°. If one ship is directly behind the other, find distance between the two ships.
Solution:
Let top of the light house A, and foot B. Points of first ship C and second ship D.
Height of light house AB = 150 m
Let the distance between light house and first ship BC = x m
Let the distance between two ships CD = y m
Angle of depression at first ship ∠XAC = ∠ACB = 45°

In ΔACB, ∠B = 90°
tan∠ACB = $$\frac{\text { Opposite side }}{\text { Adjacent side }}$$ = $$\frac{\mathrm{AB}}{\mathrm{BC}}$$
tan 45° = $$\frac{150}{x}$$
1 = $$\frac{150}{x}$$ (∵ tan 45° = 1)
∴ x = 150 m

In ΔABD, ∠B = 90°
tan∠ACB = $$\frac{\mathrm{AB}}{\mathrm{BD}}$$
tan 30° = $$\frac{150}{B C+C D}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{150}{x+y}$$
x + y = 150 (∵ tan 45° = 1)
150 + y = 150 √3 (∵ x = 150)
y = 150 √3 – 150
y = 150 (√3 – 1) = 150 (1.732 – 1)
= 150 (0.732)
y = 109.5 m
Therefore distance between two ships
= 109.5 m

Question 11.
A person from the top of a building of height 15 meters observes the top and the bottom (foot) of a cell tower with the angle of elevation as 60° and the angle of depression as 45° respectively. Then find the height of that cell tower.
Solution:

Angle of elevation of the top of the tower = 60°.
Angle of depression to the foot of the tower = 45°.
Height of the building = 15 m.
Distance between tower and building = x m
Let, height of the tower = y m
From the figure
tan 45° = $$\frac{15}{x}$$ ⇒ 1 = $$\frac{15}{x}$$ ⇒ x = 15 m
Also tan 60° = $$\frac{y-x}{15}$$ ⇒ √3 = $$\frac{y-15}{15}$$ ⇒ 15√3 = y – 15
⇒ y = 15 + 15√3 = 15(√3 + 1)
= 15(1.732 +1)
= 15 × 2.732 = 40.98 m.

Question 12.
From the top of a tower of 50 m high, Neha observes the angles of depression of the top and foot of another building to be 45° and 60° respectively. Find the height of the building.
Solution:
Let CD = height of the building = h
AB = height of the tower = 50 m
BD = EC = distance between tower and building = d m

tan 45° = $$\frac{50-h}{d}$$ ⇒ 1 = $$\frac{50-h}{d}$$
⇒ d = 50 – h _____ (1)
tan 60° = $$\frac{50}{\mathrm{~d}}$$ ⇒ √3 = $$\frac{50}{\mathrm{~d}}$$ ⇒ $$\frac{50}{\sqrt{3}}$$
$$\frac{50}{\sqrt{3}}$$ 50 – h (∵ From (1))
50 = 50√3 – h√3
h√3 = 50√3 – 50
h = $$\frac{50 \sqrt{3}-50}{\sqrt{3}}$$ = $$\frac{50 \sqrt{3}-1}{\sqrt{3}}$$ m

Question 13.
The angle of elevation of the top of a hill from the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower high. Find the height of the hill.
Solution:

Given height of the tower =AB = 50 m
Let height of hill be CD = h m
and distance between their feet be AC = x m
∠ACB = 30°, ∠CAD = 60°
From right angled Δ ABC, tan 30° = $$\frac{\mathrm{AB}}{\mathrm{AC}}$$
$$\frac{1}{3}$$ = $$\frac{50}{x}$$
∴ x = 50√3 m.
From right angled Δ ACD, tan 60° = $$\frac{\mathrm{CD}}{\mathrm{AC}}$$
√3 = $$\frac{h}{x}$$
h = x√3 m
h = 50√3.√3 (∵ x = 50√3 m)
h = 50 × 3 ⇒ h = 150 m.

Question 14.
Two boys on either side of their school building of 20 m height observes its top at the angles of elevation 30° and 60° respectively. Find the distance between two boys.
Solution:
Let BC = Distance between two boys
AD = Height of the school building = 20 m

From ΔABD,
cot 30° = $$\frac{\mathrm{BD}}{\mathrm{AD}}$$ ⇒ √3 = $$\frac{B D}{15}$$
BD = 20√3 m.
From ΔACD,
cot 60° = $$\frac{\mathrm{DC}}{\mathrm{AD}}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{D C}{20}$$ ⇒ DC = $$\frac{20}{\sqrt{3}}$$ m
∴ BC = BD + DC
= 20√3 + $$\frac{20}{\sqrt{3}}$$ = $$\frac{80}{\sqrt{3}}$$m

Question 15.
A man observes top of a tower at an angle of elevation of 30°. When he walked 40 m towards the tower, the angle of elevation is changed to 60°. Find the height of the tower and distance from the first observation point to the tower.
Solution:

Given height of the tower we will find
BC = 40 m; CD = x m
Height of the tower = h m
From figure tan 30° = $$\frac{h}{x + 40}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{h}{x + 40}$$
x + 40 = √3h ⇒ x = √3h – 40 ____ (1)
tan 60°= $$\frac{h}{x}$$ ⇒ √3 = $$\frac{h}{\sqrt{3} h-40}$$ ∵ from(1)]
3h – 40√3 = h
2h = 40√3 ⇒ h = 20√3 m
∴ Height of the tower = 20√3 m.
x = √3h – 40 = √3 × 20√3 – 40
= 3 × 20 – 40
= 60 – 40 = 20 m
Distance from the first observation point to the tower.
BD = x + 40 = 20 + 40 = 60m.

Question 16.
A wire of length 24 m had been tied with electric pole at an angle of elevation 30° with the ground. As it is covering a long distance, it was cut and tied at angle of elevation 60° with the ground. How much length of the wire was cut ?
Solution:
Let AD be length of wire before cut = 24 m
Let AG be length of wire after cut (AC) = x m
Height of the electric pole = AB

Angle of elevation
∠BDA = 30°
∠BCA = 60°
In right Δle ABD
sin 30° = $$\frac{\mathrm{AB}}{\mathrm{AD}}$$
$$\frac{1}{2}$$ = $$\frac{A B}{24}$$ ⇒ 2AB = 24 ⇒ AB = 12 m
In right Δle ABC
sin 60° = $$\frac{\mathrm{AB}}{\mathrm{AC}}$$ ⇒ $$\frac{\sqrt{3}}{2}$$ = $$\frac{12}{\mathrm{AC}}$$
⇒ √3 AC = 24 ⇒ AC = $$\frac{24}{\sqrt{3}}$$
= 8√3 m = 8 × 1.732 = 13.856
Length of the wire was cut = 24 – 13.856 = 10.144 m

Question 17.
A man on the top of vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 seconds to change the angle of depression from 30° to 60°, then how long will the car take to reach the tower from that point.
Solution:

Position of observer = ‘D’
Initial position of car = ‘A’
then angle of depression = ∠XDA = 30°
∴ In ΔACD, ∠C = 30°
⇒ tan 30 = $$\frac{\mathrm{CD}}{\mathrm{AC}}$$ = $$\frac{1}{\sqrt{3}}$$
⇒ CD = $$\frac{A C}{\sqrt{2}}$$ ______ (1)
and after 12 seconds of time, position of car = ‘B’
then angle of depression = ∠XDB = 60° = ∠DBC
∴ In ΔBCD, ∠B = 60°
⇒ tan 60° = $$\frac{\mathrm{CD}}{\mathrm{BC}}$$ = √3
⇒ CD = BC√3 ______ (2)
∴ (1) = (2)
⇒ CD = $$\frac{A C}{\sqrt{2}}$$ = BC√3 ⇒ AC = 3BC
Now from the figure
AC = AB + BC
⇒ 3BC = AB + BC (∵ AC = 3BC)
⇒ AB = 2BC
So the time taken to cover the distance $$\overline{\mathrm{AB}}$$ or the distance 2$$\overline{\mathrm{BC}}$$ = 12 seconds
∴ Time taken to cover $$\overline{\mathrm{BC}}$$ = 6 sec (∵ $$\frac{12}{2}$$)
Then the time taken to approach the tower = time taken to cover the distance
$$\overline{\mathrm{AC}}$$ = 3$$\overline{\mathrm{BC}}$$ = 3(6) = 18 seconds
that means it takes 6 more seconds (18 – 12) to reach tower.

Question 18.
Two boys on either side of a temple of 60 m height observe its top at the angles of elevation 60° and 30°. Find the distance between the two boys.
Solution:
Height of the temple BD = 60 metres.
Angle of elevation of one person ∠BAD = 60°
Angle of elevation of another person ∠BCD = 30°
Let the distance between the first person and the temple, AD = x and distance between the second person and the temple, CD = d

tan 60° = $$\frac{\mathrm{BD}}{\mathrm{AB}}$$
√3 = $$\frac{60}{x}$$
x = $$\frac{60}{\sqrt{3}}$$ ………… (1)

From ΔBCD
tan 30° = $$\frac{\mathrm{BD}}{\mathrm{d}}$$
$$\frac{1}{\sqrt{3}}$$ = $$\frac{60}{d}$$
d = 60√3 ………… (2)
From (1) and (2) distance between the persons = BC + BA = x + d
= $$\frac{60}{\sqrt{3}}$$ + 60√3 = $$\frac{60+180}{\sqrt{3}}$$
= $$\frac{240}{\sqrt{3}}$$ = 80√3 meters.

AP 10th Class Maths Important Questions Chapter 9 Applications of Trigonometry

Question 1.
A boy observed the top of an electric pole at an angle of elevation of 30°, when the observation point is 10 meters away from the foot of the pole. Draw suitable diagram for the above situation.
Solution:

AB = Height of the Electric Pole
AC = Distance between the foot of the pole to the observer = 10 m
Angle of Elevation = 30°

Question 2.
Draw a diagram to find the height of the kite in the situation given below. “A person is flying a kite at an angle of elevation ‘a’ and the length of thread from his hand to kite is ‘l'”.
Solution:

B is position of kite.
AB = Length of thread = ‘l’
Persons hand is at ‘A’.
∠BAC – α – Angle of elevation.

Question 3.
A flag pole 4 m tall casts a 6 in shadow. At the same time, a nearby building casts a shadow of 24. m. How tall is the building?
Solution:

∴ The height of the building = 16 cm.

Question 4.
A tower is 100√3 m high.Find the angle of elevation of its top when observed from a point 100 m away from the foot of the tower.
Solution:

Height of a tower
(AB) = 100√3 m.
Distance between foot of tower and observer point (BC) = 100m
InΔABCTan θ = $$\frac{100 \sqrt{3}}{100}$$= √3 = tan 60°
∴ θ = 60°

Question 5.
Rehman observed the top of a temple at an angle of elevation of 30°, when the observation point is 24 m. away from the foot of the temple. Find the height of the temple.
Solution:
InΔABC,

Question 6.
A flag pole 7 m tall, casts a 8 m shadow. At the same time, a nearby building casts a shadow of 32 mts. How tall is the building?
Solution:

ΔABC ~ ΔDEF;
$$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}$$ ⇒ $$\frac{7}{\mathrm{DE}}=\frac{8}{32}$$
∴ DE = 28 m

Question 7.
An observer of height 1.8 m is 13.2 in away from a palm tree. The angle of elevation of the top of tree from his eyes is 45°. What is the height of the palm tree?
Solution:
Height of observer = 1.8 m = AB
Distance from palm tree = 13.2 m = AE = BD

Angle of elevation = ∠CBD = 45°
In Δ BCD = tan 45° = $$\frac{\mathrm{CD}}{\mathrm{BD}}$$
⇒ 1 = $$\frac{\mathrm{CD}}{13.2}$$ ⇒ CD = 13.2 m
∴ Height of palm tree = CE
= CD + DE
= 13.2 m + 1.8 = 15 m.

Question 8.
Two poles of heights 6 m. and 11m. stand on a plane ground. If the dis¬tance between the feet of the poles is 12 m. find the distance between their tops.
Solution:
Given,
Height of first pole = AB = 6 m.
Height of second pole = CD = 11 m

Distance between feet of poles = AC = 12 m
Distance between the tops of the pole,
i.e., BD
So, BE = AC = 12 m ’
Similarly, AB EC 6m.
Now, DE = DC – EC = 11 – 6 – 5 m
BD2 = DE2 + BE2
= 52 + 122 = 25 + 144 – 169
∴ BD = 13 m

Question 9.
An observer flying in an aeroplane at an altitude of 900 m observes two ships in front of him, which are in the same direction at an angles of depression of 60° and 30° respectively. Find the distance between the two ships.
Solution:

Question 10.
A wire of length 24 m had been tied with electric pole at an angle of elevation 30° with the ground. As it is covering a long distance, it was cut and tied at angle of elevation 60° with the ground. How much length of the wire was cut?
Solution:

Let AD be length of wire before cut = 24 m
Let AC be length of wire after cut (AC) = x m
Height of the electric pole = AB
Angle of elevation
∠BDA = 30°
∠BCA = 60°
In right Δle ABD
sin 30° = $$\frac{\mathrm{AB}}{\mathrm{AD}}$$
$$\frac{1}{2}=\frac{\mathrm{AB}}{24}$$ ⇒ 2AB = 24 ⇒ AB = 12m
In right Δle ABC
sin 60° = $$\frac{\mathrm{AB}}{\mathrm{AC}}$$ ⇒ $$\frac{\sqrt{3}}{2}=\frac{12}{\mathrm{AC}}$$
= √3 AC = 24 ⇒ AC = $$\frac{24}{\sqrt{3}}$$
= 8√3m = 8 × 1.732 = 13.856 m
Length of the wire was cut
= 24 – 13.856 = 10.144 m

Question 11.
A man on the top of vertical tower observes a car moving at a uniform speed coming direcdy towards it. If it takes 12 seconds to change the angle of depression from 30° to 60°, then how long will the car take to reach the tower from that point.
Solution:

Position of observer = ‘D’
Initial position of car = ‘A’
then angle of depression
= ∠DA = 30°
∴ In ΔACD, ∠C = 30°
⇒ tan 30 = $$\frac{\mathrm{CD}}{\mathrm{AC}}=\frac{1}{\sqrt{3}}$$
⇒ CD = $$\frac{\mathrm{AC}}{\sqrt{3}}$$ ……………..(1)
and after 12 seconds of time, position of car = ’B’
then angle of depression
= ∠XDB = 60° = ∠DBC
∴ In ΔBCD, ∠B = 60°
⇒ tan 60° = $$\frac{\mathrm{CD}}{\mathrm{BC}}$$ = √3
⇒ CD = BC√3 ……………… (2)
∴ (1) = (2)
⇒ CD = $$\frac{\mathrm{AC}}{\sqrt{3}}$$ = BC √3 ⇒ AC = 3BC
Now from the figure
AC = AB + BC
⇒ 3BC = AB + BC ( ∵ AC = 3BC)
⇒ AB = 2BC
So the time taken to cover the distance
$$\overline{\mathrm{AB}}$$ or the distance 2$$\overline{\mathrm{BC}}$$ =12 seconds
∴ Time taken to cover $$\overline{\mathrm{BC}}$$ distance
= 6 sec (∵ $$\frac{12}{2}$$)
Then the time taken to approach the tower = time taken to cover the distance
$$\overline{\mathrm{AC}}$$ = 3$$\overline{\mathrm{BC}}$$ = 3(6) = 18 seconds that means it takes 6 more seconds (18 – 12) to reach tower.

Question 12.
Two boys on either side of a temple of 60 m height observe its top at the angles of elevation 60° and 30°. Find the distance between the two boys.
Solution:
Height of the temple BD = 60 metres.
Angle of elevation of one person ∠BAD = 60° .
Angle of elevation of another person ∠BCD = 30°
Let the distance between the first person and the temple, AD = x and distance between the second person and the temple, CD = d

Question 13.
A pole is arranged from a height of 30m from the ground, making 60° angle with earth. Then what is its length?
Solution:
Height from the earth = AB = 30m
Length of pole = BC = ?
∠ACB = 60°
Then in ΔABC AB BC

Question 14.
A long pole is broken in a storm. Top end of the broken pole touched the head of a man at a distance of ‘d\ Then find the angle between the man and the pole that before storm.
Solution:
AB – height of pole
AC – distance = d
CD = man

After storm
B’ – broken point
‘D’= ‘B’ (∵ they coincides after storm because D’ i.e., head of man which is touched by end of tree = B’)
∴ AngIe between these two points = zero.

Question 15.
A kite is flown from a building with a height (h) m with a long rope. Now the kite and the person having it are observed with angles of elevation α and β respectively by a boy. The distance between boy and building is ‘x’ m. So draw a diagram for this data.
Solution:

AB is building height = h
K is kites position.
‘C’ is position of boy.
AC = Distance between boy and building AB = X
∠DCK = Elevation angle to kite = ∝
∠BCK Elevation angle to building top = β

Question 16.
A person observes tops of two buildings with an angle of elevations 35° and 46° from the mid point in between them. So which building is higher ? Why?
Solution:
Let ‘E’ is mid point between two buildings AB and CD.
In ΔAEB let ∠E = 35°

tan 35 = $$\frac{\mathrm{AB}}{\mathrm{AE}}$$
∴ Height of building (AB)
= (AE) tan 35° ……………… (1)
In ΔECD let ∠E – 46°
∴ tan 46 = $$\frac{\mathrm{CD}}{\mathrm{EC}}$$
∴ Height of building (CD)
= EC tan 46° ……………. (2)
In (1) and (2) AE = EC
( ∵ E is mid point)
Then value of tan 35 is less than tan 46°. So, height of CD (= EC tan 46) is higher than height of AB (= EC tan 35)
(∵ tan 46 > tan 35)
(∵ values of tan increases from 0 to 90)

Question 17.
A 15m long pole forms 53 m long shadow at 8 AM in the morning. Then find the angle made by sun rays with earth.
Solution:

As shown in figure
Height of pole (AB) = 15m
Length of shadow (BC) = 5√3
∠ACB = θ
∴ In ΔABC tan θ = $$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{15}{5 \sqrt{3}}$$ = √3
∴ tan θ = √3 ⇒ θ = 60°
∴ The sun rays made 60° angle on the earth at that time.

Question 18.
A right circular cylindrical tower, height ‘h’ and radius ‘r’ stands on the ground. Let ‘P’ be a point in the horizontal plane ground and ABC be the semi-circular edge of the top of the tower such that B is the point in it nearest to P. The angles of elevation of the points A and B are 45° and 60°
respectively. Show that $$\frac{\mathbf{h}}{\mathbf{r}}$$ = $$\frac{\sqrt{3}(1+\sqrt{5})}{2}$$
Solution:
As shown in the figure

OA = BD = h (height of cylinder)
OD = r (radius of cylinder) .
‘O’ is the centre.
ABC is the edge of semicircle (in the top of cylinder).
B is nearer to the point P. So B should be at the outer edge of diameter. That means just above ‘D’.
∠DPB = 60°, ∠OPA = 45° (given)
In ΔBDP tan P = $$\frac{\mathrm{BD}}{\mathrm{DP}}$$ (P = 60°, BD = h)
⇒ √3 = $$\frac{\mathrm{h}}{\mathrm{DP}}$$ ⇒ h = DP√3 …………….(1)
In ΔAB tan P = $$\frac{\mathrm{OA}}{\mathrm{OP}}$$
(here P = 45°, OA = h)
⇒ tan 45° = $$\frac{\mathrm{OA}}{\mathrm{OP}}$$ = 1
⇒ OA = OP
So OA = h = OP = OD + DP = r + DP
So h = r + DP …………….. (2)
From the equation (1) & (2)
h = DP√3 , h = r + DP
∴ DP√3 = r + DP
So r = DP√3 – DP
r = DP (√3 – 1) ……………. (3)

## AP 10th Class English 9th Lesson Important Questions

These AP 10th Class English Important Questions 9th Lesson will help students prepare well for the exams.

## AP Board 10th Class English 9th Lesson Important Questions and Answers

Study Skills

I. Read the pie chart carefully.

1) What percentage of readers prefer Mystery books?
35%

2) How many readers favour Romance books?
20%

3) What is the percentage of readers who prefer Science Fiction books? (A)
A) 15%
B) 25%
C) 10%
A) 15%

4) If there are 400 readers in total, how many readers prefer Fantasy books? (A)
A) 100
B) 75
C) 150
A) 100

5) What proportion of readers favour Historical Fiction books? (A)
A) 5%
B) 10%
C) 15%
A) 5%

II. Read the bar chart carefully.

1) How many students passed the Math subject?
25 students

2) What is the total number of students who passed all subjects?
100 students (25 + 30 + 20 + 15 + 10)

3) In which subject did the fewest students pass? (C)
A) Science
B) English
C) Geography
C) Geography

4) How many more students passed Science than Geography? (B)
A) 15
B) 20
C) 25
B) 20

5) How many students passed either English or History? (B)
A) 30
B) 35
C) 40
B) 35

III. Study the following table carefully.

1) What is the total sales of Product A in all regions combined?
450

2) In which region did Product B have the highest sales?
North

3) Which product had the lowest sales overall? (B)
A) Product A
B) Product B
C) Product C
B) Product B

4) What is the total sales in the South region? (A)
A) 310
B) 320
C) 330
A) 310

5) What is the average sales per product in the East region? (C)
A) 90
B) 110
C) 116.7
C) 116.7

IV. Read the pie chart carefully.

1) What percentage of friends prefer Pepperoni as their pizza topping?
40%

2) How many friends prefer Hawaiian pizza?
10%

3) What is the percentage of friends who prefer Margherita pizza? (A)
A) 20%
B) 25%
C) 15%
A) 20%

4) If there are 50 friends in total, how many friends prefer Vegetarian pizza? (A)
A) 7
B) 8
C) 10
A) 7

5) What proportion of Triends prefer Meat Feast pizza? (A)
A) 15%
B) 10%
C) 20%
A) 15%

Section – B

Grammar & Vocabulary

Combining Sentences Using who/which/that, etc.

Exercise – 1

1. The actor won an Oscar. He starred in the latest blockbuster. (Combine the sentences using ‘who’)
The actor who starred in the latest blockbuster won an Oscar.

2. The nurse works at the hospital. She saved a patient’s life yesterday. (Combine the sentences using ‘who’)
The nurse who saved a patient’s life yesterday works at the hospital.

3. My cousin is a professional athlete. He competes in international tournaments. (Combine the sentences using who’)
My cousin, who competes in international tournaments, is a professional athlete.

4. The restaurant serves Italian cuisine. It is located downtown. (Combine the sentences using ‘which’)
The restaurant, which is located downtown, serves Italian cuisine.

5. The laptop has a touchscreen display. It was released last year. (Combine the sentences using ‘that’)
The laptop that was released last year has a touchscreen display.

6. The weather forecast predicts rain. We should bring umbrellas. (Combine the sentences using ‘that’)
The weather forecast that predicts rain suggests we should bring umbrellas.

7. The professor gave a lecture. I attended it yesterday. (Combine the sentences using ‘whose’)
The professor whose lecture I attended yesterday gave a lecture.

8. I met a journalist. I admire her investigative reporting. (Combine the sentences using ‘whom’)
I met a journalist whom I admire for her investigative reporting.

9. We hired a gardener. I recommended him for the job. (Combine the sentences using ‘whom’)
We hired a gardener whom I recommended for the job.

10. She introduced me to her brother. I have heard great things about him. (Combine the sentences using ‘whom’)
She introduced me to her brother whom I have heard great things about.

Exercise – 2

1. The volunteer helps at the shelter. She walks the dogs every evening. (Combine the sentences using ‘who’)
The volunteer who walks the dogs every evening helps at the shelter.

2. The journalist reported on the political scandal. She works for a major newspaper. (Combine the sentences using ‘who’)
The journalist who works for a major newspaper reported on the political scandal.

3. I visited a city. It is known for its historic landmarks. (Combine the sentences using ‘that’)
I visited a city that is known for its historic landmarks.

4. They change animal skins into leather. It is called tanning. (Combine the sentences using ‘which’)
They change animal skins into leather which is called tanning.

5. The movie has an intriguing plot. It kept me on the edge of my seat. (Combine the sentences using ‘that’)
The movie that kept me on the edge of my seat has an intriguing plot.

6. Sudheer read a newspaper article. It was inspirational to him. (Combine the sentences using ‘which’)
Sudheer read a newspaper article which was inspirational to him.

7. The dog barks loudly. It annoys the neighbours. (Combine the sentences using ‘that’)
The dog that annoys the neighbours barks loudly.

8. The phone rang in the middle of the night. It woke me up. (Combine the sentences using ‘that’)
The phone that woke me up rang in the middle of the night.

9. Alekhya is a precocious girl. She is very sharp. (Combine the sentences using ‘who’)
Alekhya, who is very sharp, is a precocious girl.

10. I have a friend. I trust her with my deepest secrets. (Combine the sentences using ‘whom’)
I have a friend whom I trust with my deepest secrets.

Voice

Change the following sentences into Passive Voice.

Exercise – 1

1. The children built a sandcastle on the beach.
A sandcastle was built on the beach by the children.

2. The company launched a new advertising campaign.
A new advertising campaign was launched by the company.

3. Sarah watered the plants in the garden.
The plants in the garden were watered by Sarah.

4. The chef cooked a gourmet meal for the guests.
A gourmet meal for the guests was cooked by the chef.

5. The teacher graded the students assignments.
The students assignments were graded by the teacher.

6. John fixed the broken bicycle in the garage.
The broken bicycle in the garage was fixed by John.

7. The volunteers cleaned up the litter in the park.
The litter in the park was cleaned up by the volunteers.

8. The mechanic repaired the engine of the car.
The engine of the car was repaired by the mechanic.

9. The artist painted a beautiful portrait of his wife.
A beautiful portrait of his wife was painted by the artist.

10. The doctor treated the patient’s illness with antibiotics.
The patient’s illness was treated with antibiotics by the doctor.

Exercise – 2

1. They marketed our products.
Our products were marketed (by them).

2. They declared emergency.
Emergency was declared.

3. The teacher asked me to go and sit on the back bench.
I was asked by the teacher to go and sit on the back bench.

4. Who wrote the Ramayana?
By whom was the Ramayana written?

5. I know her.
She is known to me.

6. Someone has stolen my pen.
My pen has been stolen.

7. The police have arrested the terrorist.
The terrorist has been arrested.

8. She is writing an examination.
An examination is being written by her.

9. Sheela can sing songs.
Songs can be sung by Sheela.

10. Close the door.
Let the door be closed.

Exercise – 1

1. She forgot her keys. She couldn’t enter the house. (Combine the sentences using ‘because’)
She couldn’t enter the house because she forgot her keys.

2. He was very tired. He fell asleep during the movie. (Combine the sentences using ‘because’)
He fell asleep during the movie because he was very tired.

3. The children were noisy. I couldn’t concentrate on my work. (Combine the sentences using ‘because’)
I couldn’t concentrate on my work because the children were noisy.

4. The restaurant was full. We had to wait for a table. (Combine the sentences using ‘since’)
We had to wait for a table since the restaurant was full.

5. He had never worn shoes. His feet developed cracks. (Combine the sentences using ‘since’)
His feet developed cracks since he had never worn shoes.

6. She was nervous. She had to give a presentation. (Combine the sentences using ‘although’)
She had to give a presentation although she was nervous.

7. The students were attentive. The teacher was pleased. (Combine the sentences using ‘as’)
The teacher was pleased as the students were attentive.

8. He was late. He missed the beginning of the concert. (Combine the sentences using ‘becaqse’)
He missed the beginning of the concert because he was late.

9. The road was icy. The car skidded off the road. (Combine the sentences using ‘because’)

10. She was busy. She forgot to reply to his message. (Combine the sentences using ‘because’)
She forgot to reply to his message because she was busy.

Exercise – 2

1. She was hungry. She ate a sandwich. (Combine the sentences using ‘because’)
She ate a sandwich because she was hungry.

2. The weather was cold. They decided to stay indoors. (Combine the sentences using ‘because’)
They decided to stay indoors because the weather was cold.

3. He missed the train. He arrived late for work. (Combine the sentences using ‘since’)
He arrived late for work since he missed the train.

4. The music was loud. They couldn’t hear each other. (Combine the sentences using ‘because’)
They couldn’t hear each other because the music was loud.

5. The room was small. They managed to fit all the furniture. (Combine the sentences using ‘although’)
They managed to fit all the furniture although the room was small.

6. He was ill. He went to the doctor. (Combine the sentences using ‘because’)
He went to the doctor because he was ill.

7. The movie was boring. They left before it finished. (Combine the sentences using ‘because’)
They left before the movie finished because it was boring.

8. She was excited. She couldn’t sit still. (Combine the sentences using ‘because’)
She couldn’t sit still because she was excited.

9. The traffic was heavy. He arrived late for the meeting. (Combine the sentences using ‘when’)
He arrived late for the meeting when the traffic was heavy.

10. The cat was playful. It chased its tail around the house. (Combine the sentences using ‘as’)
The cat chased its tail around the house as it was playful.

Suitable Prepositions

Fill in the blanks with suitable prepositions given in the brackets.

Exercise – 1

1. The project was delayed the unexpected weather conditions. (in front of/ because of/ according to)
2. the schedule, the meeting will start at 9:00 AM sharp. (In addition to / According to / For the sake of)
3. She made the reservation her family members. (on behalf of / due to / by means of)
4. his age, he managed to complete the marathon. (In front of/ In spite of/ For the sake of)
5. The team succeeded their hard work and dedication. (in front of/ because of/ for the sake of)
6. The conference was attended by delegates from many countries the world. (due to / in addition to / according to)
7. He apologized the inconvenience caused. (in front of/ due to / for the sake of)
8. She was awarded the scholarship her academic achievements. (in addition to / on behalf of / because of)
9. the instructions, please submit your application by the deadline. (According to / Because of/ In front of)
10. The project was completed successfully the team’s collaborative efforts. (in front of/ by means of/ because of)
1. because of
2 According to
3. on behalf of
4. In spite of
5. because of
7. due to
8. because of
9. According to
10. because of

Exercise – 2

1. The ball landed the roof. (in / on / at)
2. They went for aj walk the beach. (in / on / at)
3. The cookies are the jar. (in / on / at)
4. The cat is the box. (in /on / at)
5. We arrived the airport. (in /on /at)
6. The painting is the easel. (in / on / at)
7. The spider spun its web the corner. (in /on /at)
8. The keys were left the counter. (in / on / at)
9. She sat the edge of the bed. (in / on / at)
10. The boat sailed the river. (in / on / at)
1. on
2. on
3. in
4. in
5. at
6. on
7. in
8. on
9. on
10. on

Suitable Forms of the Verbs

Fill in the blanks in the following sentences using correct form of the verb given in brackets.

Exercise – 1

1. By the time I arrived, they ____ (finish) their lunch.
2. She ____ (work) as a teacher for ten years before she retired.
3. They ____ (not/go) to the cinema last night because they had already seen the movie.
4. By next year, I ____ (save) enough money to buy a new car.
5. He ____ (live) in London for five years before he moved to Paris.
6. When ____ I was a child, I ____ (dream) of becoming a pilot.
7. She ____ (cook) dinner when the guests arrived.
8. By the time they arrive, I ____ (clean) the house.
9. They ____ (travel) to Japan twice in the past.
10. I ____ (not/see) that movie yet.
3. did not go
4. will have saved
6. used to dream
7. was cooking
8. will have cleaned
9. travelled
10. have not seen

Exercise – 2

1. The team ____ (is/are) planning a strategy for the game.
2. One of the students ____ (has/have) forgotten to submit the assignment.
3. The pile of books on the table ____ (needs/need) to be organized.
4. Mathematics ____ (is/are) his favorite subject.
5. The news about the accident ____ (is/are) disturbing.
6. The committee members _____ (has/have) decided on the new policy.
7. A pair of shoes ____ (was/were) left in the hallway.
8. The herd of elephants ____ (was/were) grazing in the field.
9. The majority of the students ____ (prefer/prefers) to study in groups.
10. Each of the apples ____ (is/are) ripe.
1. is
2. has
3. needs
4. is
5. is
6. have
7. was
8. were
9. prefer
10. is

Giving a Suitable Advise or Suggestion

Exercise – 1

1. Your classmate is having difficulty concentrating during study sessions. Advise him / her to find a quiet and distraction-free environment, take regular breaks, and practice mindfulness or meditation techniques.
Finding a quiet and distraction-free environment, taking regular breaks, and practicing mindfulness or meditation techniques can help improve concentration during study sessions.

2. Your coworker is struggling to handle conflicts with colleagues. Advise him / her to listen actively, communicate assertively, and seek common ground for resolution.
Listening actively, communicating assertively, and seeking common ground for resolution can help handle conflicts with colleagues more effectively.

3. Your neighbour is experiencing frequent headaches and fatigue. Advise him / her to .stay hydrated, practice good posture, and consider seeking medical advice if symptoms persist.
Staying hydrated, practicing good posture, and considering seeking medical advice if symptoms persist can help alleviate headaches and fatigue.

4. Your friend wants to become more environmentally conscious but doesn’t know where to start. Advise him / her to reduce, reuse, and recycle, conserve energy, and educate themselves on sustainable practices.
Reducing, reusing, and recycling, conserving energy, and educating yourself on sustainable practices can help you become more environmentally conscious.

5. Your sibling is struggling to find motivation for regular exercise. Advise him / her to choose activities they enjoy, set achievable goals, and find an exercise buddy for accountability.
Choosing activities you enjoy, setting achievable goals, and finding an exercise buddy for accountability can help you find motivation for regular exercise.

6. Your teammate wants to improve their leadership skills. Advise him / her to lead by example, communicate effectively, and seek feedback from team members.

7. Your roommate is having trouble sticking to a healthy diet. Advise him / her to plan meals ahead, keep healthy snacks readily available, and practice mindful eating habits.

8. Your colleague is struggling with public speaking anxiety. Advise him / her to practice relaxation techniques, visualize success, and start with smaller speaking engagements to build confidence.
Practicing relaxation techniques, visualizing success, and starting with smaller speaking engagements can help alleviate public speaking anxiety.

9. Your family member wants to improve their financial literacy. Advise him / her to educate themselves on budgeting, saving, and investing, and seek advice from financial professionals if needed.
Educating yourself on budgeting, saving, and investing, and seeking advice from financial professionals can help improve your financial literacy.

10. Your friend is feeling overwhelmed by social media pressure. Advise him / her to limit screen time, curate their social media feeds, and prioritize real-life connections and activities.
Limiting screen time, curating social media feeds, and prioritizing real-life connections and activities can help alleviate social media pressure.

Exercise – 2

1. Your classmate is struggling to stay focused while studying. Advise him / her to eliminate distractions, break study sessions into shorter intervals, and practice active learning techniques.
Eliminating distractions, breaking study sessions into shorter intervals, and practicing active learning techniques can help improve focus while studying.

3. Your neighbour is having difficulty coping with loss. Advise him / her to seek support from friends and family, consider joining a support group, and allow themselves time to grieve.
Seeking support from friends and family, considering joining a support group, and allowing yourself time to grieve can help cope with loss.

4. Your friend wants to improve their time management skills. Advise him / her to use a planner or digital calendar, set realistic goals, and practice saying no to non-essential tasks.
Using a planner or digital calendar, setting realistic goals, and practicing saying no to non-essential tasks can help improve time management skills.

5. Your sibling is struggling to find a job after graduation. Advise him / her to network with professionals in their field, tailor their resume to each job application, and consider seeking internships or volunteer opportunities for experience.
Networking with professionals, tailoring their resume, and considering internships or volunteer opportunities can help you find a job.

6. Your teammate is experiencing writer’s block while working on a project. Advise him / her to take a break and engage in a different creative activity, brainstorm ideas with a colleague, and consider changing their environment for inspiration.
Taking a break and engaging in a different creative activity, brainstorming ideas with a colleague, and changing their environment for inspiration can help overcome writer’s block.

7. Your roommate is feeling stressed about managing finances. Advise him / her to create a budget, track expenses, and explore ways to increase income or reduce spending.
Creating a budget, tracking expenses, and exploring ways to increase income or reduce spending can help manage finances more effectively.

8. Your colleague is struggling to maintain a work-life balance. Advise him / her to set boundaries between work and personal life, schedule leisure activities, and prioritize self-care.
Setting boundaries, scheduling leisure activities, and prioritizing self-care can help maintain a healthy work-life balance.

9. Your family member is experiencing difficulty sleeping at night. Advise him / her to establish a bedtime routine, create a comfortable sleep environment, and limit screen time before bed.
Establishing a bedtime routine, creating a comfortable sleep environment, and limiting screen time before bed can help Improve sleep quality.

10. Your friend is feeling anxious about an upcoming presentation. Advise him / her to practice relaxation techniques, visualize success, and focus on the content rather than perfection.
Practicing relaxation techniques, visualizing success, and focusing on the content can help alleviate anxiety about an upcoming presentation.

Changing a Sentence into a Polite Request

Change the following sentence into a polite request.

Exercise – 1

Would you mind lending me your umbrella?

2. You to your coworker: “Email me the presentation slides.”
Could you please email me the presentation slides?

3. You to your sibling: “Set the table for dinner.”
Could you please set the table for dinner?

4. You to your neighbour: “Return my lawnmower when you’re done.”
Would you mind returning my lawnmower when you’re done?

6. You to your roommate: “Take out the trash.”
Could you please take out the trash?

7. You to your partner: “Make reservations for dinner tonight.”
Would you mind making reservations for dinner tonight?

8. You to your teacher: “Review my essay and provide feedback.”
Would you mind reviewing my essay and providing feedback?

9. You to your boss: “Consider me for the leadership training programme.”
Would you mind considering me for the leadership training programme?

10. You to your parents: “Pick me up from practice.”
Could you please pick me up from practice?

Exercise – 2

Would you mind lending me your textbook?

2. You to your coworker: “Print this document for me.”
Could you please print this document for me?

3. You to your sibling: “Pick up some milk on your way home.”

4. You to your neighbour: “Keep your dog from barking late at night.”
Would you mind keeping your dog from barking late at night?

5. You to your classmate: “Explain this math problem to me.”
Would you mind explaining this math problem to me?

6. You to your roommate: “Turn down the volume on the TV.”
Would you mind turning down the volume on the TV?

7. You to your partner: “Make reservations for our anniversary dinner.”
Would you mind making reservations for our anniversary dinner?

9. You to your boss: “Provide me with more training opportunities.”
Would you mind providing me with more training opportunities?

10. You to your parent: “Drop me off at the bus station.”
Could you please drop me off at the bus station?

Identifying the Expression

What do the following sentences mean? Put a tick (✓) mark against the right answer.

Exercise – 1

1. i) Could you please lend me a hand with this heavy box?
A) Requesting help (✓)
B) Refusing help ( )
C) Seeking permission ( )

ii) I apologize for the inconvenience caused by the delay.
A) Apologizing (✓)
B) Offering help ( )
C) Requesting ( )
D) Complimenting ( )

2. i) m not sure which movie to watch tonight. Any recommendations?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission ( )

A) Apologizing ( )
B) Requesting (✓)
C) Ordering ( )
D) Thanking ( )

3. i) I’m having trouble figuring out how to solve this math problem. Can you help me?
A) Offering help (✓)
B) Refusing help ( )
C) Seeking permission ( )
D) Seeking help ( )

ii) I’m sorry, but I cannot accompany you to the party tonight.
A) Apologizing ( )
B) Refusing help ( )
C) Declining invitation (✓)

4. i) Do you mind if I sit here?
A) Offering help ( )
B) Refusing permission ( )
C) Seeking permlission (✓)
D) Offering help ( )

ii) Please remember to water the plants while I’m away.
A) Apologizing ( )
B) Requesting (✓)
C) Ordering ( )
D) Refusing permission ( )

5. i) I’m not sure if I should buy this dress or not. What do you think?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission ( )

ii) I’m sorry if my remarks came across as insensitive.
A) Apologizing (✓)
B) Refusing permission ( )
C) Thanking ( )
D) Declining invitation ( )

Exercise – 2

1. i) If you need assistance with your homework, just let me know.
A) Offering help (✓)
B) Refusing help ( )
C) Seeking permission ( )

ii) You must wear your seatbelt while driving.
A) Apologizing ( )
B) Refusing permission ( )
C) Requesting ( )
D) Ordering (✓)

2. i) I’m having a hard time making a decision. Can you offer any guidance?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission ( )

ii) I’m sorry, but I won’t be able to lend you my car this weekend.
A) Apologizing ( )
B) Offering help ( )
C) Declining invitation ( )
D) Refusing permission (✓)

3. i) No, thank you. I can manage by myself.
A) Offering help ( )
B) Refusing help (✓)
C) Seeking permission ( )

ii) Would you mind helping me with this heavy box?
A) Apologizing ( )
B) Requesting (✓)
C) Complimenting ( )
D) Ordering ( )

4. i) Is it alright if I leave early today?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission (✓)

ii) I’m really sorry, but I cannot join you for lunch today.
A) Apologizing ( )
B) Offering help ( )
C) Declining invitation (✓)
D) Ordering ( )

5. i) Let me know if you need any assistance with the project.
A) Offering help (✓)
B) Refusing help ( )
C) Seeking permission ( )

ii) Please ensure that you complete the report by tomorrow morning.
A) Apologizing ( )
B) Requesting (✓)
C) Refusing permission ( )
D) Ordering ( )

Synonyms

Read the paragraph and write the Synonyms of the underlined words choosing the , words given in the box.

1. [agreement, calculated, shaking, insane, piled, heartbeats]
LOMOV : [greatly moved] Honoured Stepan Stepanovitch, do you think I may count on her consent (a)?

CHUBUKOV : Why, of course, my darling, and… as if she won’t consent! She’s in love; egad, she’s like a lovesick cat, and so on. Shan’t be long! [Exit.]

LOMOV : It’s cold… I’m trembling (b) all over, just as if I’d got an examination before me. The great thing is, I must have my mind made up. If I give myself time to think, to hesitate, to talk a lot, to look for an ideal, or for real love, then I’ll never get married. Brr… It’s cold! Natalya Stepanovna is an excellent housekeeper, not bad-looking, well-educated. What more do I want? But I’m getting a noise in my ears from excitement. [Drinks] And it’s impossible for me not to marry. In the first place, I’m already 35 – a critical age, so to speak. In the second place, I ought to lead a quiet and regular life. I suffer from palpitations (c), I’m excitable and always getting awfully upset; at this very moment my lips are trembling, and there’s a twitch in my right eyebrow. But the very worst of all is the way I sleep. I no sooner get into bed and begin to go off, when suddenly something in my left side gives a pull, and I can feel it in my shoulder and head… I jump up like a lunatic (d), walk about a bit and lie down again, but as soon as I begin to get off to sleep there’s another pull! And this may happen twenty times… [Natalya Stepanovna comes in.]
(a) agreement
(b) shaking
(c) heartbeats
(d)insane

2. [stuck, farmers, disposition, swamp, eEternity, press]

LOMOV : How? I’m speaking of those Oxen Meadows which are wedged (a) in between vour birchwoods and the Burnt Marsh (b).

NATALYA : Yes, yes… they’re ours.

LOMOV : No, you’re mistaken, honoured Natalya Stepanovna, they’re mine.

NATALYA : Just think, Ivan Vassilevitch! How long have they been yours?

LOMOV : How long? As long as I can remember.

NATALYA : Really, you won’t get me to believe that!

LOMOV : But you can see from the documents, honoured Natalya Stepanovna. Oxen Meadows, it’s true, were once the subject of dispute, but now everybody knows that they are mine. There’s nothing to argue about. You see my aunt’s grandmother gave the free use of these Meadows in perpetuity (c) to the peasants of your father’s grandfather, in return for which they were to make bricks for her. The peasants (d) belonging to your father’s grandfather had the free use of the Meadows for forty years, and had got into the habit of regarding them as their own, when it happened that…
(a) stuck
(b) swamp
(c) eternity
(d) farmers

3. [injustice, fields, calculated, currency, meddle, pulsating]
NATALYA : No, it isn’t at all like that! Both grandfather and great- grandfather reckoned (a) that their land extended to Burnt Marsh – which means that Oxen Meadows (b) were ours. I don’t see what there is to argue about. It’s simply silly!

LOMOV : I’ll show you the documents, Natalya Stepanovna!

NATALYA : No, you’re simply joking, or making fun of me. What a surprise! We’ve had the land for nearly three hundred years, and then we’re suddenly told that it isn’t ours! Ivan Vassilevitch, I can hardly believe my own ears. These Meadows aren’t worth much to me. They only come to five dessiatins, and are worth perhaps 300 roubles (c), but I can’t stand unfairness (d). Say what you will, I can’t stand unfairness.
(a) calculated
(b) fields
(c) currency
(d) injustice

4. [nightdress, piled, bewildered, magnificent, laborers, lineage]
NATALYA : You must excuse my apron and neglige (a). We’re shelling peas for drying. Why haven’t you been here for such a long time? Sit down… [They seat themselves.] Won’t you have some lunch?

NATALYA : Then smoke. Here are the matches. The weather is splendid (b) now, but yesterday it was so wet that the workmen (c) didn’t do anything all day. How much hay have you stacked (d) ? Just think, 1 felt greedy and had a whole field cut, and now I’m not at all pleased about it because I’m afraid my hay may rot. I ought to have waited a bit. But what’s this? Why, you’re in evening dress! Well, I never! Are you going to a ball or what? Though I must say you look better… Tell me, why are you got up like that?
(a) nightdress
(b) magnificent
(c) laborers
(d) piled

5. [heartbeats, meddle, plot, meddle, rudeness, purebred]
CHUBUKOV : Yes really, what sort of a hunter are you, anyway? You ought to sit at home with your palpitations (a), and not go tracking animals. You could go hunting, but you only go to argue with people and interfere (b) with their dogs and so on. Let’s change the subject in case I lose my temper (c). You’re not a hunter at all, anyway!

LOMOV : And are you a hunter? You only go hunting to get in with the Count and to intrigue (d). Oh, my heart! You’re an intriguer!
(a) heartbeats
(b) meddle
(c) mood
(d) plot

6. [terrible, stumbles, pulsating, recognize, quiet, control]
NATALYA : Why talk rot? It’s awful (a)! It’s time your Guess was shot, and you compare him with Squeezer!

LOMOV : Excuse me, I cannot continue this discussion, my heart is palpitating (b).

NATALYA : I’ve noticed that those hunters argue most who know least.

LOMOV : Madam, please be silent (c). My heart is going to pieces.
[shouts] Shut up!

NATALYA : I shan’t shut up until you acknowledge (d) that Squeezer is a hundred times better than your Guess!
(a) terrible
(b) pulsating
(c) quiet
(d) recognize

7. [stumbles, lineage, strawman, bewildered, rudeness, nomads]
[Lomov staggers (a) out.]
CHUBUKOV : Devil take him!

NATALYA : What a rascal! What trust can one have in one’s neighbours after that!

CHUBUKOV : The villain! The scarecrow (b)!

NATALYA : The monster! First he takes our land and then he has the impudence (c) to abuse us.

CHUBUKOV : And that blind hen, yes, that turnip-ghost has the confounded (d) cheek to make a proposal, and so on! What? A proposal!
(a) stumbles
(b) strawman
(c) rudeness
(d) bewildered

8. [exhausted, purebred, agonizing, exceeded, lineage, insanity]
NATALYA : Of course he’s better! Of course, Squeezer is young, he may develop a bit, but on points and pedigree (a) he’s better than anything that even Volchanetsky has got.

LOMOV : Excuse me, Natalya Stepanovna, but you forget that he is overshot (b), and an overshot always means the dog is a bad hunter!

NATALYA : Overshot, is he? The first time I hear it!

LOMOV : I assure you that his lower jaw is shorter than the upper.

NATALYA : Have you measured?

LOMOV : Yes. He’s all right at following, of course, but if you want to get hold of anything…

NATALYA : In the first place, our Squeezer is a thoroughbred (c) animal, the son of Harness and Chisels while there’s no getting at the pedigree of your dog at all. He’s old and as ugly as a worn-out (d) cab-horse.
(a) lineage
(b) exceeded
(c) purebred
(d) exhausted

9. [beating, decanter, nomads, harvesters, torturous, curved]
NATALYA : I can make you a present of them myself, because they’re mine! Your behaviour, Ivan Vassilevitch, is strange, to say the least! Up to this we have always thought of you as a good neighbour, a friend; last year we lent you our threshing-machine, although on that account we had to put off our own threshing (a) till November, but you behave to us as if we were gypsies (b). Giving me my own land, indeed! No, really, that’s not at all neighbourly! In my opinion, it’s even impudent, if you want to know.

LOMOV : Then you make out that I’m a landgrabber? Madam, never in my life have I grabbed anybody else’s land and I shan’t allow anybody to accuse me of having done so. [Quickly steps to the carafe (c) and drinks more water] Oxen Meadows are mine!

NATALYA : It’s not true, they’re ours!

LOMOV : Mine!

NATALYA : It’s not true! I’ll prove it! I’ll send my mowers (d) out to the Meadows this very day!
(a) beating
(c) decanter
(d) harvesters

10. [rough, heart flutter, control, agonizing, alcoholic, swilling]
LOMOV : [Clutches at his heart] Oxen Meadows are mine! You understand? Mine!

NATALYA : Please don’t shout! You can shout yourself hoarse (a) in your own house but here I must ask you to restrain (b) yourself!

LOMOV : If it wasn’t, madam, for this awful, excruciating (c) palpitation (d), if my whole inside wasn’t upset, I’d talk to you in a different way! [Yells] Oxen Meadows are mine!
(a) rough
(b) control
(c) agonizing
(d) heart flutter

11. [madness, alcoholic, drinking, hunched, surpassed, flailing]
LOMOV : Never mind about my people! The Lomovs have all been honourable people, and not one has ever been tried for embezzlement, like your grandfather!

CHUBUKOV : You Lomovs have had lunacy (a) in your family, all of you!

NATALYA : All, all, all!

CHUBUKOV : Your grandfather was a drunkard (b), and your younger aunt, Nastasya Mihailovna, ran away with an architect, and so on…

LOMOV : And your mother was hump-backed (c). [Clutches at his heart] Something pulling in my side… My head…. Help! Water!

CHUBUKOV : Your father was a guzzling (d) gambler!
(b) alcoholic
(c) hunched
(d) drinking

Antonyms

Read the paragraph and match the words given in Column ‘A’ with the Antonyms in Column ‘B’.

1. We just get along somehow, my angel (a), thanks to your prayers, and so on. Sit down, please do… Now, you know, you shouldn’t forget (b) all about your neighbours, my darling. My dear fellow, why are you so formal (c) in your get-up! Evening dress, gloves, and so on. Can you be going anywhere (d), my treasure?

 A B a) angel 1) nowhere b) forget 2) burden c) formal 3) ignored d) anywhere 4) informal 5) remember 6) devil

a-6, b-5, c-4, d-1

2. Well, you see, it’s like this. [Takes his arm] I’ve come to you, honoured (a) Stepan Stepanovitch, to trouble (b) you with a request (c). Not once or twice have I already had the privilege (d) of applying to you for help, and you have always, so to speak… I must ask your pardon, I am getting excited. I shall drink some water, honoured Stepan Stepanovitch.

 A B a) honoured 1) disadvantage b) trouble 2) demand c) request 3) recall d) privilege 4) ease 5) disrespected 6) relaxed

a-5, b-4, c-2, d-1

3. My dear fellow… I’m so glad, and so on… Yes, indeed, and all that sort of thing. [Embraces and kisses Lomov] I’ve been hoping (a) for it for a long time. It’s been my continual desire (b). [Sheds a tear] And I’ve always loved you, my angel, as if you were my own son. May God give you both – His help and His love and so on, and so much hope… What am I behaving in this idiotic (c) way for? I’m off my balance (d) with joy, absolutely off my balance! Oh, with all my soul… I’ll go and call Natasha, and all that.

 A B a) hoping 1) intelligent b) desire 2) imbalance c) idiotic 3) aversion d) balance 4) despairing 5) disrespected 6) relaxed

a-4, b-3, c-1, d-2

4. It’s cold… I’m trembling (a) all over, just as if I’d got an examination before me. The great thing is, I must have my mind made up. If I give myself time to think, to hesitate (b) , to talk a lot, to look for an ideal, or for real love, then I’ll never get married. Brr… It’s cold! Natalya Stepanovna is an excellent housekeeper, not bad-looking, well- educated. What more do I want? But I’m getting a noise in my ears from excitement (c). [Drinks] And it’s impossible for me not to marry. In the first place, I’m already 35 – a critical age, so to speak. In the second place, I ought to lead a quiet and regular life. I suffer from palpitations, I’m excitable and always getting awfully upset; at this very moment my lips are trembling, and there’s a twitch in my right eyebrow. But the very worst (d) of all is the way I sleep.

 A B a) trembling 1) decide b) hesitate 2) comfort c) excitement 3) steady d) worst 4) best 5) burden 6) boredom

a-3, b-1, c-6, d-4

5. Then smoke. Here are the matches. The weather is splendid (a) now, hut yesterday it was so wet that the workmen didn’t do anything all day. How much hay have you stacked? Just think, I felt greedy (b) and had a whole field cut, and now I’m not at all pleased (c) about it because I’m afraid my hay may rot. I ought to have waited a bit. But what’s this? Why, you’re in evening dress! Well, I never! Are you going to a ball or what? Though I must say you look better (d)… Tell me, why are you got up like that?

 A B a) splendid 1) displeased b) greedy 2) worse c) pleased 3) apathy d) better 4) firm 5) selfless 6) dull

a-6, b-5, c-1, d-2

6. I shall try to be brief. You must know, honoured Natalya Stepanovna, that I have long, since my childhood, in fact, had the privilege of knowing your family. My late aunt and her husband, from whom, as you know, I inherited (a) my land, always had the greatest respect (b) for your father and your late mother. The Lomovs and the Chubukovs have always had the most friendly (c), and I might almost say the most affectionate (d), regard for each other. And, as you know, my land is a near neighbour of yours. You will remember that my Oxen Meadows touch your birchwoods.

 A B a) inherited 1) aloof b) respect 2) generous c) friendly 3) ordinary d) affectionate 4) unfriendly 5) disrespect 6) acquired

a-6, b-5, c-4, d-1

7. But you can see ifrom the documents, honoured Natalya Stepanovna. Oxen Meadows, it’s true (a) were once the subject of dispute (b) but now everybody knows that they are mine. There’s nothing to argue (c), about. You see my aunt’s grandmother gave the free use of these Meadows in perpetuity (d) to the peasants of vour father’s grandfather, in return for which they were to make bricks for her.

 A B a) true 1) impermanence b) dispute 2) departure c) argue 3) earned d) perpetuity 4) agree 5) harmony 6) false

a-6, b-5, c-4, d-1

8. No, you’re simply joking (a), or making fun of me. What a surprise! We’ve had the land for nearly three hundred years. and then we’re suddenly (b) told that it isn’t ours! Ivan Vassilevitch. I can hardly believe (c) my own ears. These Meadows aren’t worth much to me. They only come to five dessiatins, and are worth perhaps 300 roubles, but I can’t stand unfairness (d). Say what you will. I can’t stand unfairness.

 A B a) joking 1) doubt b) suddenly 2) fairness c) believe 3) gradually d) unfairness 4) serious 5) concur 6) alienation

a-4, b-3, c-1, d-2

9. I can make you a present of them myself, because they’re mine! Your behaviour, Ivan Vassilevitch, is strange (a), to say the least (b) ! Up to this we have always thought of you as a good neighbour, a friend; last year we lent you our threshing-machine, although on that account we had to put off our own threshing till November, but you behave to us as if we were gypsies. Giving me my own land, indeed! No, really (c), that’s not at all neighbourly! In my opinion, it’s even impudent (d), if you want to know.

 A B a) strange 1) most b) least 2) solemn c) really 3) familiar d) impudent 4) polite 5) equity 6) insincerely

a-3, b-1, c-6, d-4

10. Rut, please, Stepan Stepanovitch. how can they be yours? Do be a reasonable (a) man! My aunt’s grandmother gave the Meadows for the temporary (b) and free use of your grandfather’s peasants. The peasants used the land for forty years and got accustomed (c) to it as if it was their own (d), when it happened that…

 A B a) reasonable 1) borrowed b) temporary 2) unreasonable c) accustomed 3) courteous d) own 4) justice 5) unaccustomed 6) permanent

a-2, b-6, c-5, d-1

11. Dear one, why yell like that? You won’t prove anything just by yelling (a). I don’t want anything of yours, and don’t intend (b) to give up what I have. Why should I? And you know, my beloved (c), that if you propose to go on arguing (d) about it. I’d much sooner give up the Meadows to the peasants than to you. There!

 A B a) yelling 1) concurring b) intend 2) greatest c) beloved 3) respectful d) arguing 4) hated 5) unplanned 6) whispering

a-6, b-5, c-4, d-1

12. There’s some demon of contradiction (a) in you today, Ivan Vassilevitch. First you pretend (b) that the Meadows are yours; now, that Guess is better than Squeezer. I don’t like people who don’t say what they mean, because you know perfectly (c) well that Squeezer is a hundred times better (d) than your silly Guess. Why do you want to say he isn’t?

 A B a) contradiction 1) worse b) pretend 2) agreement c) perfectly 3) unintentional d) better 4) genuine 5) imperfectly 6) harmonizing

a-2, b-4, c-5, d-1

13. Don’t excite yourself. my precious (a) one. Allow me. Your Guess certainly has his good points. He’s purebred, firm on his feet, has well-sprung ribs, and all that. But, my dear man. if you want to know the truth (b) that dog has two defects (c) he’s old and he’s short in the muzzle. It’s not true! Mv dear fellow. I’m very liable to lose my temper (d), and so, just because of that, let’s stop arguing.

 A B a) precious 1) perfection b) truth 2) quiet c) defect 3) lie d) temper 4) worthless 5) calmness 6) despised

a-4, b-3, c-1, d-5

Right Forms of the Words

Fill in the blanks in the following passage choosing the right form of the words given in brackets.

1. The play is about the tendency of _______ (a) (wealth / wealthy / wealthier) families to seek ties with other wealthy families, to increase their estates by _______ (b) (encourage / encouraged / encouraging) marriages that make good economic sense. Ivan Lomov, a long time wealthy neighbour of Stepan Chubukov, also wealthy, comes to seek the hand of Chubukov’s twenty-five-year-old daughter, Natalya. All three are quarrelsome people, and they quarrel over petty issues. The proposal is in danger of being forgotten amidst all this _______ (c) (quarrell / quarrelled / quarrelling). But economic good sense ensures that the proposal is made, after all – although the quarrelling perhaps _______ (d) (continue / continues / continued)!
a) wealthy
b) encouraging
c) quarrelling
d) continues

2. My dear fellow… I’m so glad, and so on… Yes, indeed, and all that sort of thing. [Embraces and kisses Lomov] I’ve been _______ (a) (hope / hoped / hoping) for it for a long time. It’s been my continual _______ (b) (desire / desired / desiring). [Sheds a tear] And I’ve always loved you, my angel, as if you were my own son. May God give you both – His help and His love and so on, and so much hope… What am I _______ (c) (behave / behaved / behaving) in this idiotic way for? I’m off my _______ (d) (balance/ balanced / balancing) with joy, absolutely off my balance! Oh, with all my soul… I’ll go and call Natasha, and all that.
a) hoping
b) desire
c) behaving
d) balance

3. It’s cold… I’m _______ (a) (tremble / trembled / trembling) all over, just as if I’d got an _______ (b) (examining / examination / examined) before me. The great thing is, I must have my mind made up. If I give myself time to think, to hesitate, to talk a lot, to look for an ideal, or for real love, then I’ll never get married. Brr… It’s cold! Natalya Stepanovna is an excellent housekeeper, not bad-looking, well-educated. What more do I want? But I’m getting a noise in my ears from _______ (c) (excite / excitement / excited). [Drinks] And it’s impossible for me not to marry. In the first place, I’m already 35 – a critical age, so to speak. In the second place, I ought to lead a quiet and regular life. I suffer from _______ (d) (palpitation / palpitations / palpitated), I’m excitable and always getting awfully upset; at this very moment my lips are trembling, and there’s a twitch in my right eyebrow. But the very worst of all is the way I sleep.
a) trembling
b) examination
c) excitement
d) palpitations

4. Then smoke. Here are the matches. The weather is splendid now, but yesterday it was so wet that the workmen didn’t do anything all day. How much hay have you _______ (a) (stack /stacked / stacking)? Just think, I felt greedy and had a whole field cut, and now I’m not at all _______ (b) (please / pleased /pleasing) about it because I’m afraid my hay may rot. I ought to have _______ (c) (wait / waited / waiting) a bit. But what’s this? Why, you’re in evening dress! Well, I never! Are you going to a ball or what? Though 1 must say you look _______ (d) (good / better / best)… Tell me, why are you got up like that?
a) stacked
c) waited
d) better

5. I shall try to be brief. You must know, _______ (a) (honour / honoured / honouring) Natalya Stepanovna, that I have long, since my childhood, in fact, had the _______ (b) (privilege / privileged / privileging) of knowing your family. My late aunt and her husband, from whom, as you know, I _______ (c) (inherit / inherited / inheriting) my land, always had the _______ (d) (great / greater / greatest) respect for your father and your late mother.
a) honoured
b) privilege
c) inherited
d) greatest

Vowel Clusters

Complete the spelling of the words choosing ‘ae’, ‘ea\ ‘ee’, ‘ei’, ‘oo’, ‘ou’, ‘ai\ ‘ia’/ie’, io’, oi’, iii’ or au’.
Exercise – 1

1. My d_ _r (a) fellow, whom do I s_ _(b)! Ivan Vassilevitch! I am extremely glad!
2. Sit down, pl_ _se (a) do… Now, you know, you shouldn’t forget all about your neighb_ _rs (b), my darling.
3. Can you be going anywhere, my tr_ _sure?
4. No. I’ve come only to see you, hon_ _red Stepan Stepanovitch.
5. My dear fellow… I’m so glad, and so on… Yes, ind_ _d, and all that sort of thing.
6. May God give you both – His help and His love and so on, and so much hope… What am I behaving in this id_ _tic way for?
7. Why, of c_ _rse, my darling, and… as if she won’t consent! She’s in love; egad, she’s like a lovesick cat, and so on. Shan’t be long!
8. You must excuse my apron and neglige. We’re shelling p s for drying.
9. Why haven’t you b_ _n (a) here for such a long time? Sit down… [They s_ _t (b) themselves.] Won’t you have some lunch?
10. How much hay have you stacked? Just think, I felt gr_ _dy (a) and had a whole f_ _ld (b) cut, and now I’m not at all pleased about it because I’m afraid my hay may rot.
1. (a) deer (b) see
3. treasure
4. honoured
5. indeed
6. idiotic
7. course
8. peas
9. (a) been (b) seat
10. (a) greedy (b) field

Exercise – 2

1. I shall try to be br_ _f (a). You must know, honoured Natalya Stepanovna, that I have long, since my childh_ _d (b), in fact, had the privilege of knowing your family.
2. You will remember that my Oxen M_ _dows (a) t_ _ch (b) your birchwoods.
3. How ? I’m sp_ _king (a) of those Oxen Meadows which are wedged in between your birchw_ _ds (b)and the Burnt Marsh.
4. You see my aunt’s grandmother gave the free use of these Meadows in perpet_ _ty (a) to the p_ _sants (b) of your father’s grandfather, in return for which they were to make bricks for her.
5. I don’t see what there is to arg_ _about. It’s simply silly!
6. Say what you will, I can’t stand unf_ _mess.
7. Hear me out, I implore you! The p_ _sants (a) of your father’s grandfather, as I have already had the honour of expl_ _ning (b) to you, used to bake bricks for my aunt’s grandmother.
8. I can’t make head or t_ _l (a) of all this about nts (b) and grandfathers and grandmothers.
9. Ours! You can go on proving it for two days on end, you can go and put on fift_ _n dress jackets, but I tell you they’re ours, ours, ours!
10. Ivan Vassilevitch, is strange, to say the least! Up to this we have always thought of you as a good n_ _ghbour (a), a friend; last year we lent you our threshing-machine, although on that acc_ _nt (b) we had to put off our own threshing till November, but you behave to us as if we were gyps_ _s (c).
1. (a) brief (b) childhood
3. (a) speaking (b) brichwoods
4. (a) perpetuity (b) peasants
5. argue
6. unfairness
7. (a) peasants (b) explaining
8. (a) tail (b) aunts
9. fifteen
10. (a) neighbour (b) account (c) gypsies

Suffixes or Inflections

Fill in the blanks of the following words with correct suffixes.

Exercise – 1

1. Then you make out that I’m a landgrabber? Madam, never in my life have I grabb___(ing / ed) anybody else’s land and I shan’t allow anybody to accuse me of having done so.
2. If it wasn’t, madam, for this awful, excruciating palpitat___(a) (ed / ion), if my whole inside wasn’t upset, I’d talk to you in a differ___(b) (ant / ent) way! [Yells] Oxen Meadows are mine!
3. What’s the matter? What are you shout___(ed / ing) for?
4. But, please, Stejpan Stepanovitch, how can they be yours? Do be a reason___(ing / able) man!
5. Dear one, why yell like that? You won’t prove anything just by yell___(ing / ed).
6. Why should I? And you know, my belov___(a) (ed / ing), that if you propose to go on argu___(b) (ment / ing) about it, I’d much sooner give up the Meadows to the peasants than to you. There!
7. I, young man, am twice your age, and ask you to speak to me without agitat___(ed / ing) yourself, and all that.
8. No, you just think I’m a fool and want to have me on! You call my land yours, and then you want me to talk to you calm___(a) (ness / ly) and polite___(b) (ness / ly)!
9. The Lomovs have all been honour___(a) (ed / able) people, and not one has ever been tried for embezzle___(b) (ness / ment), like your grandfather!
10. Your father was a guzzl___(ed / ing) gambler!
1. grabbed
2. (a) palpitation (b) different
3. shouting
4. reasonable
5. yelling
6. (a) beloved (b) arguing
7. agitating
8. (a) calmly (b) politely
9. (a) honourable (b) embezzlement
10. guzzling

Exercise – 2

1. And that blind hen, yes, that turnip-ghost has the confound___(a) (ing / ed) cheek to make a propos___(b) (el / al), and so on! What? A proposal!
2. [Lomov enters, exhausted.] My heart’s palpitat___(a) (ed / ing) awful___(b) (ly / ness). My foot’s gone to sleep. There’s something that keeps pulling in my side….
3. Forgive us, Ivan Vassilevitch, we were all a little heat (ing / ed).
4. My heart’s beating awfully. My Meadows… My eyebrows are both twitch___(ed / ing)…..
5. Are you going to start shoot___(ed / ing) soon?
6. I’m thinking of having a go at the blackcock, honour___(ing / ed) Natalya Stepanovna, after the harvest.
7. I don’t know. Must have got his leg twist___(ed / ing) or bitten by some other dog.
8. I assure you that his lower jaw is short___(a) (er / est) than the upper. Have you measur___(b) (ment / ed)?
9. Yes. He’s all right at follow___(ed / ing), of course, but if you want to get hold of anything…
10. I don’t like people who don’t say what they mean, because you know perfect___(ion / ly) well that Squeezer is a hundred times better than your silly Guess.
1. (a) confounded (b) proposal
2. (a) palpitating (b) awfully
3. heated
4. twitching
5. shooting
6. honoured
7. twisted
8. (a) shorter (b) measured
9. following
10. perfectly

Identifying the Wrongly Spelt Word

Find the wrongly spelt word and write the correct spelling.

Exercise – 1

1. reception
2. procession
3. quarrelsome
4. amidst
5. treasure
6. assistant
7. interrupt
8. privilege
9. affectionate
10. peasants

Exercise – 2

1. roubles
2. accustomed
3. honourable
4.sausage
5. proposal
6. lunacy
7. machine
8. acknowledge
9. interfere
10. temporary

Dictionary Skills

1. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘affectionate’?

b) What is the synonym of the word ‘affectionate’?
loving

2. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘awfully’?

b) What is the synonym of the word ‘awfully’?
terribly

3. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the synonym of the word ‘continual’?
continuous

b) What is the adverb form of the word ‘continual’?
continually

4. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘excruciating’?

b) Write any two collocations using the word ‘excruciatingly’.
excruciatingly painful; excruciatingly boring

5. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the noun form of the word ‘handsome’?
handsomeness

b) What are the other degrees of comparison of the word ‘handsome’?
handsome, handsomer, handsomest / handsome, more handsome, most handsome

6. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘ideal’?

b) Write the adverb form of the word ‘ideal’.
ideally

7. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What are the synonyms of the word ‘malicious’?
malevolent, spiteful

b) What is the adverb form of the word ‘malicious’?
maliciously

8. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘precious’?

b) What are the synonyms of the word ‘precious’?
treasured, affected

9. Read the following dictionary entry of the word.

Now, answer the following questions using the information above
a) What is the part of speech of the word ‘privilege’?
noun, verb

b) What is the synonym of the word ‘privilege’?
honour

10. Read the following dictionary entry of the word.

Now, answer the following questions using the information above
a) What is the part of speech of the word ‘splendid’?

b) Write any two collocations using the word ‘splendid’.
splendid scenery; splendid idea

11. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the pasft form of the word ‘upset’?
upset

b) What is the synonym of the word ‘upset’?
distress

12. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the antonym of the word ‘willing’?
unwilling

b) What is the adverb form of the word ‘willing’?
willingly

Classification of Words

Arrange the following words under correct headings.

1. [merchant, meadows, housekeeper, lunatic, birchwood, Burnt Marsh, peasant, field]

 Persons Places 1. 1. 2. 2. 3. 3. 4. 4.

 Persons Places 1. merchant 1. meadows 2. housekeeper 2. birchwood 3. lunatic 3. Burnt Marsh 4. peasant 4. field

2. [neighbour, gypsy, aunt, grandfather, land grabber, intriguer, husband, grandmother]

 Persons Relations 1. 1. 2. 2. 3. 3. 4. 4.

 Persons Relations 1. neighbour 1. aunt 2. gypsy 2. grandfather 3. land grabber 3. husband 4. intriguer 4. grandmother

3. [gloves, peas, apron, knife, pistol, neglige, jacket, carafe]

 Articles of clothing Things 1. 1. 2. 2. 3. 3. 4. 4.

 Articles of clothing Things 1. gloves 1. peas 2. apron 2. knife 3. neglige 3. pistol 4. jacket 4. carafe

4. [fox, sheep, bullet, mowers, machine, dog, horse, hay]

 Animals Things 1. 1. 2. 2. 3. 3. 4. 4.

 Animals Things 1. fox 1. bullet 2. sheep 2. machine 3. dog 3. mowers 4. horse 4. hay

Matchings

Match the phrases given under Columan ‘A’ with their meanings given under Column ‘B’.
1.

 Column – A Column – B 1) brrr A) expresses surprise, anger or affirmation 2) egad B) an expression meant to ward off the evil eye 3) my word C) used to express hesitation 4) tfoo D) an exclamation used to suggest shivering E) used to show that you are surprised at some-thing F) used to express your happiness

1-D, 2-A, 3-E, 4-B

2.

 Column – A Column – B 1) hear me out A) an object made to resemble a human figure 2) on principle B) rude and not showing any respect 3) to say the least C) a way to grab someone’s attention 4) has gone to sleep D) to act in accordance with one’s moral ideals E) not to describe something in the strongest way you could F) a feeling of numbness or tinging known temporary paresthesia

1-C, 2-D, 3-E, 4-F

3.

 Column – A Column – B 1) pettifogger A) used to expess discomfort, aversion, or impertinence 2) scarecrow B) an inferior legal practitioner 3) milk sop C) an object made to resemble a human figure set up to scare birds away from a field 4) ouf D) to give a short jerk or convulsive movement E) a person who is indecisive and lacks in courage F) to beg somebody earnestly

1-B, 2-C, 3-E, 4-A

4.

 Column – A Column – B 1) my feet A) used to express, discomfort or aversion 2) verst B) a way to grab one’s attention 3) lunatic C) used to mean that you do not believe what another person has told you 4) I’ll cut my throat D) a person who does crazy things that are often dangerous E) Russian measure of 1.1 kilometre F) to behave in a way that will cause harm to yourself

1-C, 2-E, 3-D, 4-F

5.

 Column – A Column – B 1) my heart is biurst A) to start 2) worrying a sheep B) to make a decision 3) lose my temper C) to give way from an excess of emotion 4) has gone to sleep D) chases and frightens it might also bite it E) to fail / manage to control your anger F) become numb

1-C, 2-D, 3-E, 4-F

6.

 Column – A Column – B 1) you are under the slipper of your house-keeper A) you are scared of your housekeeper 2) to make up one’s mind B) explode 3) can’t make the head or tail of all C) to make a decision 4) go off D) unable to understand E) start F) rude and not showing any respect

1-A, 2-C, 3-D, 4-B

SECTION – C

CREATIVE EXPRESSION

Conversation

Question 1.
When Lomov came to the Chubukov’s house to propose to Natalya’s hand in marriage, Natalya and her father Stepanovitch Chubukov abused Lomov, made derogatory and defaming remarks against Lomov and his family. They assaulted him and made a scathing attack on him and his family. Later they pitied him and they were sorry for that attack against him. They talked to each other and they accused themselves of being rude and indecent against Lomov.
What would be the conversation between Natalya and her father Chubukov in this context?
Chubukov : I’m very sorry for the young man ! I pity him.
Natalya : Me too. We are very much rude, impolite and indecent towards him.
Chubukov : Yes. We cursed him, abused him and drove him out. I am very sorry my dear, Lomov.
Natalya : We made derogarory remarks against him. We made defaming remarks against his family members. I too beg his pardon.
Chubukov : It’s all because of you.
Natalya : No, it was because of you!
Chubukov : I was very much delighted and thrilled when he made the proposal of having your hand in marriage.
Natalya : You should have told me about it before it all happened.
Chubukov : I thought, he would make the proposal to you straight away. But you’ve taken the conversation to another direction.
Natalya : No. I was not. He was obstinate and adamant in his arguments. He claimed that Oxen Meadows belonged to him.
Chubukov : You too did the same. You got aggravated and made a heated argument with him by being contemptuous and arrogant and inexorable.
Natalya : You also humiliated him, made a scathing attack on his personality.
Chubukov : I’m very sorry for him. Please pardon my dear sir. We’ve defamed you with our arrogance and we’ve behaved brutally towards you.
Natalya : Sir Lomov ! I too pity you. We are inexcusable for our outrageous attitude towards you. I beg your pardon. We have hurt you with our savage, barbaric and unscrupulous outburst towards you.
Chubukov : He is actually polite, amiable and social.
Natalya : Yes. He is a gentleman but he is nervous and excitable. He is very much sensitive and is suffering from palpitations because he is distraught, anxious and perturbed with our excoriated attack on him.
Chubukov : We should not have behaved so brutally against him. Our attitude towards him is inexpiable whereas he is affable.
Natalya : Dad ! I implore you ! 1 beg you earnestly to fetch him back. I’d like to marry him. First I want to apologise to him.
Chubukov : I will bring him provided you make a promise that you should not make any illogical arguments against him.
Natalya : I assure you, I won’t make any petty arguments against him.
Chubukov : Somehow, I’ll persuade him, coax him and try to appease him with my pleasing words. See you.

Question 2.
Environmental conservation focuses on the protection, preservation, and sustainable management of natural resources and ecosystems. It encompasses efforts to safeguard biodiversity, mitigate pollution, address climate change, and promote overall environmental sustainability.
You have read the lesson ‘The Proposal’ and understood the significance of marshes and meadows in ecosystems and the importance of preserving wetlands. Now write a possible conversation between you and your science teacher.
I : Good morning, Mr. Smith. I wanted to talk to you about something I learned in our literature class that I think is relevant to our science curriculum.
Teacher : Good morning! Of course, I’m always interested in connecting different subjects. What did you learn?
I : Well, we were discussing the play “The Proposal,” and there was a part about the importance of marshes and meadows in ecosystems. It got me thinking about environmental conservation and how crucial it is to preserve wetlands.
Teacher : That’s an excellent connection to make! Wetlands are incredibly important ecosystems that provide numerous benefits to both wildlife and humans. They act as natural filters, improving water quality, and they serve as habitats for diverse plant and animal species.
I : Yes, exactly! I learned that wetlands also help prevent flooding by absorbing excess water during heavy rainfall and storms. Plus, they store carbon, which helps mitigate climate change.
Teacher : Absolutely. Wetlands are among the most productive ecosystems on Earth, supporting a wide array of species and providing valuable services to the environment and society. It’s crucial that we work to preserve and protect them.
I : I completely agree. I think it’s important for us to raise awareness about the significance of wetlands and advocate for their conservation. Maybe we could even organize a school project or field trip to a nearby wetland area.
Teacher : That’s a fantastic idea! Engaging in hands-on activities and getting out into nature is an excellent way to foster appreciation and understanding of the environment. Let’s discuss it further and see how we can incorporate it into our curriculum.
I : Great! I’m really passionate§bout environmental conservation, and I think it’s essential for us to take action to protect our planet’s precious ecosystems.
Teacher : I couldn’t agree more. It’s inspiring to see students like you who are enthusiastic about making a positive difference. Let’s work together to make a meaningful impact on the preservation of wetlands and the environment as a whole.

Question 3.
You have read the lesson The Book That Saved The Earth and understood the theme of miscommunication and misunderstanding in the play and its real-life implications.
Now write a possible conversation between you and your friend Kiran.
You : Hey Kiran, have you ever heard of the play “The Book That Saved The Earth”?
Kiran : No, I haven’t. What’s it about?
You : It’s a hilarious story set in the future where Martians attempt to invade Earth but end up misinterpreting nursery rhymes as threats. It’s all about miscommunication and misunderstanding.
Kiran : That sounds intriguing! So, how does miscommunication play out in the story?
You : Well, the Martians, led by their egotistical leader Think-Tank, misinterpret innocent nursery rhymes as coded messages indicating Earth’s military capabilities. For example, they interpret “Mistress Mary, quite contrary, How does your garden grow? With cockle shells and silver bells” as evidence of Earthlings growing explosives and rare metals in their gardens.
Kiran : That’s hilarious! But it also highlights how misinterpretation can lead to serious consequences. Have you ever been in a situation where miscommunication caused problems?
You : Definitely! I remember once during a group project, there was a misunderstanding about deadlines which led to confusion and unnecessary stress. Thankfully, we were able to sort it out by openly discussing our concerns and clarifying expectations.
Kiran : It’s amazing how miscommunication can escalate if not addressed promptly. The play seems like a humorous way to shed light on such a common issue.
You : Absolutely! It’s a lighthearted reminder of the importance of clear communication and understanding, both in fictional scenarios and real life.
Kiran : I’ll definitely have to check out the play sometime. Thanks for telling me about it!
You : No problem! I think you’ll enjoy it. Plus, it’s a great conversation starter about the complexities of communication and the consequences of misunderstanding.

Diary Entry

Question 1.
When Lomov came to Chubukov’s house to propose to Natalya’s hand in marriage, Natalya and her father Chubukov abused Lomov, made derogatory and defaming remarks against him. Moreover she assaulted him and made a scathing attack against him and his family. He was hurt and went away humiliated. Natalya repented after retrospecting her brutal, savage, unscrupulous outburst towards him. She wanted to write her feelings in a diary entry.
Assume yourself to be Natalya and write your feelings in this context as a diary entry.
4 December, 20xx
Monday
9.30 p.m.
Dear Diary,
Natalya

Question 2.
You have read the lesson The Book That Saved the Earth and seen the portrayal of aliens and extraterrestrial life in literature and media.
Write a Diary Entry in about 100 words.
March 12, 20xx
Dear Diary,
Today was an intriguing day as I delved into the captivating world of literature, particularly exploring the portrayal of aliens and extraterrestrial life in the play “The Book That Saved the Earth.” It was a fascinating journey through the imaginative realm of science fiction, where Martians, led by the egotistical Think-Tank, attempted to invade Earth but were thwarted by a simple misunderstanding of nursery rhymes.

As I reflected on the play’s portrayal of aliens, I couldn’t help but marvel at the creativity and diversity of interpretations that writers bring to the concept of extraterrestrial beings. From benevolent visitors to menacing invaders, aliens have been depicted in various forms across literature and media, reflecting humanity’s curiosity and imagination about the unknown.

What struck me most about the play was its underlying theme of miscommunication and misunderstanding. Despite their advanced technology and intellect, the Martians failed to grasp the true meaning behind Earth’s cultural artifacts, leading to humorous yet consequential misinterpretations. It made me ponder how miscommunication can occur even between beings of different species, highlighting the importance of empathy, curiosity, and open-mindedness in fostering mutual understanding.

As I closed the book, I found myself contemplating the broader implications of miscommunication in our own lives. How often do we misinterpret others’ intentions or messages due to our own biases or lack of understanding? How can we strive to bridge the gap and foster genuine connections with those who are different from us?

Tonight, as I drift off to sleep, I carry with me a newfound appreciation for the power of storytelling to not only entertain but also provoke thought and reflection on the complexities of human nature and the vast mysteries of the universe.
Until next time,
xxxx

Letter Writing

Question 1.
Thanking a mentor for their wisdom, encouragement, and belief in one’s potential fosters a deep sense of appreciation and respect. Recognizing the mentor’s rol6 not only strengthens the mentor-mentee bond but also inspires others to pay it forward in mentoring relationships.
Pen a letter to a mentor expressing your gratitude for their guidance.
[City, State, ZIP Code]
[Phone Number]
[Date]
[Mentor’s Name]
[Mentor’s Position/Organization]
[City, State, ZIP Code]
Dear [Mentor’s Name],
I hope this letter finds you well. As I reflect on my journey of growth and development, I am filled with an overwhelming sense of gratitude for the profound impact you have had on my life.’ I wanted to take a moment to express my deepest appreciation for your unwavering guidance, encouragement, and belief in my potential.

When I first embarked on this journey, I was filled with uncertainty and apprehension, unsure of my abilities and unsure of what lay ahead. It was during this pivotal moment that you graciously stepped into my life, offering me not only your expertise and wisdom but also your unwavering support and encouragement.

Your mentorship has been nothing shdrt of transformative, empowering me to embrace challenges, navigate obstacles, and strive for excellence in every endeavor. Your insights, advice, and encouragement have not only helped me achieve my goals but have also inspired me to reach for heights I never thought possible.

More than just imparting knowledge and skills, you have instilled in me a sense of confidence, resilience, and determination to pursue my passions and aspirations fearlessly. Your belief in my potential has served as a beacon of light during moments of doubt and uncertainty, reminding me of the boundless possibilities that lie within reach.

I am profoundly grateful for the countless hours you have dedicated to nurturing my growth and development, patiently guiding me through challenges, celebrating my successes, and offering words of wisdom and encouragement along the way. Your mentorship has truly been a gift, and 1 am honored to have had the opportunity to learn from someone as wise, compassionate, and inspiring as you.

As I continue on my journey, I carry with me the invaluable lessons and insights you have imparted, knowing that your guidance will continue to shape and inspire me in all that I do. I am deeply humbled by your belief in me, and I am committed to honoring your legacy by paying it forward and empowering others on their paths to success.

Thank you, from the bottom of my heart, for your unwavering support, guidance, and belief in me. Your mentorship has made all the difference, and I am eternally grateful for the profound impact you have had on my life.
With deepest gratitude and respect,

Question 2.
Climate change profoundly affects the environment. Rising temperatures lead to melting ice caps, raising sea levels and threatening coastal areas. Extreme weather events like hurricanes, droughts, and wildfires become more frequent and severe, endangering ecosystems and human lives. Shifts in rainfall patterns disrupt agriculture, impacting food security worldwide. Ocean acidification harms marine life, jeopardizing delicate ecosystems like coral reefs. Biodiversity loss accelerates as habitats become uninhabitable for many species. The impacts of climate change ripple across the planet, posing significant challenges to sustainable development, human health, and the delicate balance of nature.
Pen a letter to a friend discussing the impact of climate change on the environment.
[City, State, ZIP Code]
[Phone Number]
[Date]
[Friend’s Name]
[City, State, ZIP Code]
Dear [Friend’s Name],
I hope this letter finds you well. I wanted to reach out to you to discuss a topic that weighs heavily on my mind lately: the profound impact of climate change on our environment.
Climate change is no longer just a distant threat-it’s a reality that we are witnessing unfold before our eyes. The consequences of rising global temperatures are far-reaching and devastating, affecting ecosystems, communities, and livelihoods across the planet.

One of the most visible effects of climate change is the melting of ice caps and glaciers, leading to rising sea levels. As a result, coastal areas are increasingly threatened, putting millions of people at risk of displacement and loss of homes and livelihoods. The frequency and severity of extreme weather events like hurricanes, droughts, and wildfires are also on the rise, wreaking havoc on communities and ecosystems alike.

Shifts in rainfall patterns and prolonged droughts are disrupting agriculture and food security worldwide, leading to crop failures, food shortages, and increased prices.

These impacts of climate change are not isolated – they are interconnected and exacerbate existing social, economic, and environmental challenges. The most vulnerable communities, including marginalized populations and developing countries, bear the brunt of the consequences, despite contributing the least to global greenhouse gas emissions.

Through sustainable practices, conservation efforts, and advocacy for climate action, we can make a meaningful difference in safeguarding our planet for future geherations. I urge you to join me in raising awareness, taking action, and advocating for policies that prioritize environmental protection and climate resilience.

Together, we have the power to shape a more sustainable and equitable future for all living beings on Earth. Let’s stand united in our commitment to protect and preserve the precious gift of our planet.
With hope and determination,

Question 3.
Business owners who prioritize community well-being deserve appreciation for their valuable contributions and dedication to making a positive difference. Their dedication strengthens local economies, providing jobs and resources. Supporting community events, charities, and initiatives demonstrates a commitment to social responsibility. By fostering partnerships and investing in neighborhood improvement projects, they enhance the quality of life for residents.
Draft a letter to a local business owner praising their contribution to the community.
[City, State, ZII? Code]
[Phone Number]
[Date]
[City, State, ZIP Code]
I hope this letter finds you well. I am writing to express my sincere appreciation and gratitude for your remarkable contributions to our community. Your unwavering dedication to making a positive difference has not gone unnoticed, and I wanted to take a moment to commend you for your exemplary efforts.

As a local business owner, you play a vital role in strengthening our community and enhancing the well-being of its residents. Your commitment to social responsibility is truly commendable, and it is evident that you prioritize the needs and interests of our community in all aspects of your business endeavors.

Moreover, I am deeply impressed by your active engagement in community events, charities, and initiatives that aim to address pressing social and environmental issues. Your willingness to lend a helping hand and support worthy causes speaks volumes about your genuine commitment to the well-being of our community.

Your contributions extend far beyond the walls of your business establishment. Through your generosity and compassion, you have fostered a sense of unity and belonging among residents and have inspired others to join you in your mission to create positive change.

I want to express my heartfelt gratitude for your tireless dedication, passion, and leadership in championing community well-being. Your efforts have not only enriched the fabric of our community but have also set a shining example for others to follow.

Thank you once again for your invaluable contributions to our community. Your efforts are deeply appreciated and serve as a beacon of hope and inspiration for us all.

Biographical Sketch

Question 1.
Write a biographical sketch of Rameshbabu Praggnanandaa using the following information.
Full Name : Rameshbabu Praggnanandaa
Nick Name : Praggu
Date of Birth : 10 August 2005
Birth Place : Chennai, Tamil Nadu
Father : Rameshbabu
Mother : Nagalakshmi
Sister : Vaishali (elder)
Occupation / Profession : Chess player
Achievements : – Won the World Youth Chess Championship, 2013 at the age of seven
– Won the World Youth Championship in 2015
– became the youngest International Master in Chess in 2016
– Achieved his first grandmaster norm at the World Junior Chess Championship
– Earned his second grandmaster norm at GM Norm Tournament in Greece
– Became the fifth one who won the grandmaster title in 2022
– Lost to World No. 1 Magnus Carlsen in the FIDE Chess World Cup 2023 and became the youngest Runner-up of FIDE Championships
Rameshbabu Praggnanandaa is an Indian chess grandmaster. He is a chess prodigy. His full name is Praggnanandaa Rameshbabu. His nick name is Praggu. He was born on 10 August, 2005. He was born at Chennai, Tamil Nadu. His father is Rameshbabu. His mother’s name is Nagalakshmi. His elder sister’s name is Vaishali. He is a professional chess player. He is now only 18 years old. Though he is young he has achieved a lot. He won the World Youth Chess Championship, 2013 at the age of seven. He won the World Youth Chess Championship again in 2015. He became the youngest International Chess Master in 2016. He achieved his first grandmaster norm at the World Junior Chess Championship. He earned his second grandmaster norm at GM Norm Tournament in Greece. He lost to the World No. 1 Magnus Carlsen in the FIDE World Cup’2023 and became the youngest Runner-up of FIDE Championships.

Question 2.
Write a biographical sketch of Jagdeep Dhankhar, the Vice-President of India.
Name : Jagdeep Dhankhar
Born : May 18, 1951 in Jhunjhunu district in Rajasthan
Age : 71 years
Political party : Bharatiya Janata Party
Parents : Gokal Chand and Kesari Devi
Spouse : Sudesh Dhankhar; Daughter : Kamna
Education : School education : Sainik School, Chittorgarh B.Sc and LLB from University of Rajasthan, Jaipur.
Profession : A lawyer, designated as the Senior Advocate by the High Court of Rajasthan
Occupation : Politics
Previous offices : Governor of West Bengal; Member of the Rajasthan Legislative Assembly; Minister of State for Parliamentary Affairs; Member of . Parliament (Lok Sabha); elected as the Vice-President of India in ‘ 2022
Jagdeep Dhankhar is the present Vice-President of India. He was born on 18th May 1951 in Jhunjhunu district of Rajasthan. He is now 71 years old. His parents were Gokal Chand and Kesari Devi. His wife’s name is Sudesh Dhankhar. He had his school education at Sainik School in Chittorgarh. He studied B.Sc and LLB in University of Rajasthan, Jaipur. He was a professional lawyer. He was designated as the Senior Advocate by High Court of Rajasthan. His occupation is politics. He belonged to Bharatiya Janata Party. He held many positions. He was the Governor of West Bengal. He was the Member of Rajasthan Legislative Assembly. He was the Minister of State for Parliamentary Affairs. He was the Member of (Lok Sabha) Parliament. He was elected as the Vice-President of India in 2022.

Question 3.
Write a biographical sketch of Smt. Nigar Shaji, the Director of the first space based project Aditya L-l launched by India to study the Sun.
Name : Nigar Shaji
Nick Name : Nigar
Date of Birth : 1964
Age : 59 years
Place of Birth : Sengottai, Tamil Nadu
Father : Sheikh Meeran; Mother: Saitoon Biwi
Education : – Post graduate in Electronics and Communications from BIT, Ranchi
– Early schooling : SRM Girls School in Sengottai
– Degree in Electronics and Communications from Madurai Kamaraj University
Profession : ISRO scientist
Joined SHAAR (Satish Dhawan Space Center) – a division of ISRO in 1987
Contribution : – Her unshakable dedication and expertise brought her to UR Rao Satellite Centre in Bengaluru
– Made significant contributions to the design and construction of Indian communication remote sensing and interplanetary satellites
– She was the Associate Project Director for the Resource Sat- 2A, a remote sensing satellite
– Became the Project Director of India’s first Solar Mission Aditya L-l and ensured its successful launch from Sriharikota on September 2, 2023
Srimathi Nigar Shaji is the Director fo the first space project Aditya L-l launched by India to study the Sun. Her nick name is Nigar. She was born in 1964 and is now 59 years old. She was born at Sengottai in Tamil Nadu. Her father is Sheikh Meeran and her mother is Saitoon Biwi. She had her early schooling at SRM Girls School in Sengottai. She did her degree in Electronics and Communications from Madurai Kamaraj University. She is a post graduate in Electronics and Communications from BIT, Ranchi. She is an ISRO scientist by profession. She joined SHAAR (Satish Dhawan Space Centre) – a division of ISRO in 1987. Her unshakable dedication and expertise brought her to UR Rao Satellite Centre in Bengaluru. She made significant contributions to the design and construction of Indian communication remote sensing and interplanetary satellites. She was the Associate Project Director for the Resource Sat – 2A, a remote sensing satellite. She became the Project Director of India’s first Solar Mission Aditya L-l and ensured its successful launch from Sriharikota on September 2, 2023.

Question 4.
Write a biographical sketch of Smt. Draupadi Murmu, the 15th President of India.
Original Name : Puti Biranchi Tudu
Birth : 20 June 1958; Age : 64 years
Place of Birth : Uparbeda; Mayur Bhanj, Odisha
Tribe : Santhali tribal family
Father’s Name : Biranchi Narayan Tudu
Education : Elementary education : at local primary school at Uparbeda
Secondary education : Girls’ High School
Earlier profession : Professor
Married : Shyam Charan Murmu, a bank officer
Previous offices : Governor of Jharkhand; Minister of State for Fisheries and Animals; Minister of State for Commerce and Transport Sworn in as the 15th President of India on 25th July, 2022.
Draupadi Murmu is the 15th President of India. She is the first tribal woman elected as the President of India. Her original name is Puti Biranchi Tudu. She was born on 20 June 1958. She is now 64 years old. She was born at Uparbeda, Mayur Bhanj in Odisha. She belongs to Santali tribal family. Her father’s name was Biranchi Narayan Tudu. She had her elementary education at Uparbeda. She had her secondary education in Girls’ High School. She completed her graduation from Ramadevi Women’s College. She was a professor before she entered politics. She married Shyam Charan Murmu, a bank officer. She was the Governor of Jharkhand. She was the Minister of Fisheries and Animals. She was the Minister of State for Commerce and Transport. She was sworn in as the 15th President of India on 25th July, 2022.

Framing ‘Wh’ Questions

Read the following passage carefully focussing on the underlined parts.

1. Noodle : (bowing) 0 Great and Mighty Think-Tank (A), most powerful and intelligent creature in the whole universe, what are your orders?
Think-Tank : (peevishly) (B) You left out part of my salutation, Apprentice Noodle (C)- Go over the whole thing again.
Noodle : It shall be done, sir. (in a singsong) O Great and Mighty Think-Tank (D), Ruler of Mars and her two moons, most powerful and intelligent creature in the whole universe (E)— (out of breath) what-are-your-orders?
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) Who is the most powerful and intelligent creature?
B) How did he Think-Tank speak?
C) Who was asked to do the whole thing again?
D) Who is the ruler of Mars and her two moons?
E) What kind of Martian is the Think-Tank?

2. Think-Tank : That’s better, Noodle. I wish to be placed in communication with our manned space probe (A) to that ridiculous planet (B) we are going to put under our generous rulership. What do they call it again?
Noodle : Earth (C). your Intelligence.
Think-Tank : Earth (D) – of course. You see how insignificant the place is. But first, something important. My mirror. I wish to consult my mirror (E).
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) What does the Think-Tank wish to do?
B) What is the Earth considered as?
C) What are they going to put under their generous rulership?
D) Which is considered to be the insignificant place?
E) What does he want to consult?

3. Noodle : (speaking into a microphone) Mars Space Control calling the crew of Probe One (A). Come in, Captain Omega (B). and give us your location.
Omega : Captain Omega to Mars Space Control. Lieutenant Iota, Sergeant Oop, and I have arrived on Earth without incident (C). We have taken shelter in this (indicates room) – this square place. Have you any idea, where we are, Lieutenant Iota?
Iota : I can’t figure it out, Captain. I’ve counted two thousand of (D) of these peculiar items. This place must be some sort of storage barn (E). What do you think, Sergeant Oop?
Now Frams ‘WH’ questions to get the underlined parts in the passage as answers.
A) Who is the Mars Space Control calling?
B) Who is asked to give his location?
C) How have they arrived on Earth?
D) How many peculiar items are counted?
E) What does lota imagine the place must be?

4. Think-Tank : I understand now. Since Earth creatuers are always eating, the place in which you find yourselves is undoubtedly a crude refreshment stand (A).
Omega : (to Iota and Oop) He says we’re in a refreshment stand (B).
Oop : Well, the Earthlings (C), cetainly have a strange diet.
Think-Tank : That item in your hand is called a sandwich (D).
Omega : (nodding) A sandwich.
Iota : (nodding) A sandwich.
Oop : (taking book from his head) A sandwich?
Think-Tank : Sandwiches (E) are the main staple of Earth diet.
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) What does Think-Tank imagine the place must be?
B) What do Iota and Oop say that they are in?
C) Who have certainly a strange diet?
D) What is the item in their hand called?
E) What are the main staple of Earth diet?

5. Iota : (dubiously) Eat it? Oh, Captain! It’s a very great honour to be the first Martian to eat a sandwich (A), I’m sure, but — but how can I be so impolite as to eat before my Sergeant (B)? Sergeant (C) Oop, I order you to eat the sandwich immediately (D).
Oop : (making a face) Who, Lieutenant? Me, Lieutenant?
Iota and Omega : (saluting) For the glory of Mars, Oop!
Oop : Yes, of course! (unhappily) Immediately. (He opens his mouth wide. Omega and Iota watch him breathlessly (E)).
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) What is the great honour?
B) What is impolite?
C) What is Oop?
D) What does lota order Sergeant Oop?
E) How do lota and Omega watch him?

6. Noodle : Well, sir, I have seen surveyor films of those sandwiches (A). I noticed that the Earthlings did not eat them (B)- They used them as some sort of communication device (C).
Think-Tank : (haughtily) Naturally. That was my next point. These are actually communication sandwiches (D). Think-Tank (E) is never wrong.
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) What has he seen?
B) What did he notice?
C) What did they use them as?
D) What are they actually?
E) Who is never wrong?

7. Think-Tank : (alarmed) Stop! This is no time for levity (A)- Don’t you realise the seriousness of this discovery ? The Earthlings (B) have discovered how to combine agriculture and mining (C)- They can actually grow crops of rare metals such as silver. And cockle shells. They c*m grow high explosives (D). too. Noodle, contact our invasion fleet (E).
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) What is this not a time for?
B) Who have discovered how to combine agriculture and mining?
C) What have the Earthlings discovered?
D) What can they grow?
E) Whom do Think-Tank order Noodle to contact?

8. Historian : (chuckling) And that’s how one dusty old book of nursery rhymes (A) saved the world from a Martian invasion (B)- As you all know, in the twenty-fifth century, five hundred years after all this happened, we Earthlings (C) resumed contact with Mars, and we even became very friendly with the Martians. By that time, Great and Mighty Think-Tank had been replaced by a very clever Martian — the wise and wonderful (D) Noodle (E)!
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) What saved the world from a Martian invasion?
B) What did the dusty old book of nursery rhymes save the world from?
C) Who resumed contact with Mars?
D) What kind of a Martian was Noodle?

Information Transfer

Question 1.
Study the following bar graph carefully and write a paragraph based on the information given in it.

The bar graph illustrates the financial growth of a particular entity over a five-year period, revealing a consistent upward trend in revenue generation. Beginning at $500,000 in 2017, the entity experienced steady annual increases, reaching$600,000 in 2018, $700,000 in 2019,$800,000 in 2020, and peaking at $900,000 in 2021. This progressive expansion in revenue signifies the entity’s ability to adapt to market dynamics, capitalize on opportunities, and effectively manage resources. The substantial growth observed over the years suggests successful strategic planning, sound financial management, and potentially the introduction of innovative products or services that have resonated with consumers. The bar graph underscores the entity’s resilience and capacity for sustained growth, positioning it favorably for continued success in the future. Question 2. Study the following pie chart carefully and write a paragraph based on the information given in it. Answer: The pie chart provides insights into the preferences of travelers regarding their preferred vacation destinations. Beach resorts emerge as the most favored choice, constituting 40% of the pie, indicating the’widespread appeal of coastal getaways and seaside relaxation. Mountains follow suit, representing 25% of preferences, reflecting the allure of scenic landscapes, outdoor activities, and tranquil mountain retreats among travelers. Historical cities claim a significant portion at 20%, highlighting the enduring interest in cultural exploration, heritage sites, and immersive experiences in ancient and historically rich destinations. Theme parks secure a smaller but notable slice of the pie at 10%, showcasing the enduring popularity of amusement parks and entertainment complexes among families and thrill-seekers. Lastly, the countryside, while representing a smaller proportion at 5%, still captures the interest of travelers seeking tranquility, rural experiences, and connections with nature. The pie chart illuminates the diverse preferences and varied interests of travelers, reflecting the multitude of options available for memorable and enriching vacation experiences. Question 3. Study the following table carefully and write a paragraph based on the information given in it, Causes of Road Accidents in India  Causes of Road Accidents Number of Incidents Speeding 8500 Drunk Driving 5200 Distracted Driving 3700 Poor Road Conditions 2800 Reckless Overtaking 2500 Answer: The table presents a breakdown of the causes of road accidents in India, shedding light on the factors contributing to traffic incidents across the country. Speeding emerges as the leading cause, with a staggering 8500 incidents reported, indicating the pervasive issue of excessive speed among drivers on Indian roads. Following closely behind is drunk driving, accounting for 5200 accidents, highlighting the significant dangers posed by driving under the influence of alcohol. Distracted driving, with 3700 ’ incidents, underscores the risks associated with divided attention while operating vehicles. Poor road conditions contribute to 2800 accidents, emphasizing the importance of infrastructure maintenance and road safety measures. Reckless overtaking, responsible for 2500 accidents, highlights the need for better enforcement of traffic regulations and responsible driving practices. The table underscores the multifaceted nature of road safety challenges in India, emphasizing the urgent need for comprehensive measures to address these issues and improve traffic safety nationwide. ## AP 10th Class English 8th Lesson Important Questions These AP 10th Class English Important Questions 8th Lesson will help students prepare well for the exams. ## AP Board 10th Class English 8th Lesson Important Questions and Answers Study Skills I. Read the pie chart carefully. Now, answer the following questions. 1) What percentage of total expenses does employee salaries represent? Answer: 25% 2) How much of the total expenses is spent on marketing/advertising? Answer: 5% 3) Which expense category accounts for the smallest portion of the total expenses? (A) A) Marketing/Advertising B) Inventory/Supplies C) Rent Answer: A) Marketing/Advertising 4) If the total monthly expenses are$10,000, how much is spent on utilities? (B)
A) $1500 B)$2000
C) $2500 Answer: B)$2000

5) What is the combined percentage of expenses for inventory/supplies and miscellaneous items? (C)
A) 10%
B) 15%
C) 20%
C) 20%

II. Read the bar chart carefully.

1) How many days experienced 3 days of snowfall?
3 days

2) What is the total number of days observed for snowfall?
30 days (10 + 5 + 8 + 3 + 2 + 2)

3) Which number of days of snowfall occurred the most? (A)
A) 0 days
B) 1 day
C) 2 days
A) 0 days

4) How many more days experienced 2 days of snowfall than 4 days? (B)
A) 4
B) 6
C) 5
B) 6

5) How many days experienced at least 1 day of snowfall? (B)
A) 25
B) 20
C) 15
B) 20

III. Read the pie chart carefully.

1) What percentage of the total income is earned from online courses?
20%

2) How much of the total income is generated from affiliate marketing?
5%

3) Which income source contributes the least to the total income? (B)
A) Freelance Platforms
B) Affiliate Marketing
C) Online Courses
B) Affiliate Marketing

4) If the total monthly income is $5000, how much is earned from client projects? (B) A)$2500
B) $3000 C)$3500
B) \$3000

5) What is the combined percentage of income earned from freelance platforms and online courses? (C)
A) 15%
B) 20%
C) 35%
C) 35%

IV. Study the following table carefully.
Number of Goals Scored by Different Football Players in a League

 Player Goals Scored Ronaldo 30 Messi 28 Salah 25 Lewandowski 32 Mbappe 27

1) Who scored the most goals in the season?
Lewandowski

2) How many goals did Mbappe score?
27

3) Who scored the least number of goals? (C)
A) Ronaldo
B) Messi
C) Salah
C) Salah

4) What was the total number of goals scored by all players? (C)
A) 140
B) 1400
C) 142
C) 142

5) Who scored more goals, Messi or Salah? (A)
A) Messi
B) Salah
C) Both scored the same number of goals
A) Messi

V. Read the bar chart carefully.

1) How many students have birthdays in June?
4 students

2) What is the total number of students whose birthdays are recorded?
40 students (5 + 8+10 + 6 + 7 + 4)

3) In which month do most students have their birthdays? (B)
A) February
B) March
C) April
B) March

4) How many more students have birthdays in March than in May? (B)
A) 2
B) 3
C) 4
B) 3

5) How many students have birthdays in January and April? (A)
A) 11
B) 10
C) 12
A) 11

SECTION – B

GRAMMAR VOCABULARY

Combining Sentences Using who/which/that, etc.

Exercise -1

1. The scientist discovered a new species. She works at the research institute. (Combine the sentences using ‘who’)
The scientist who works at the research institute discovered a new species.

2. My brother is a skilled mechanic. He fixed my car last week. (Combine the sentences using ‘who’)
My brother, who fixed my car last week, is a skilled mechanic.

3. The teacher gave us a challenging assignment. She expects us to complete it by Friday. (Combine the sentences using ‘who’)
The teacher who gave us a challenging assignment expects us to complete it by Friday.

4. The house has a beautiful garden. It is located in the countryside. (Combine the sentences using ‘which’)
The house, which is located in the countryside, has a beautiful garden.

5. The smartphone has a long battery life. It was released last month. (Combine the sentences using ‘which’)
The smartphone, which was released last month, has a long battery life.

6. The movie received mixed reviews. I watched it last night. (Combine the sentences using ‘that’)
The movie that I watched last night received mixed reviews.

7. The dog is very friendly. It greeted me at the door. (Combine the sentences using ‘that’)
The dog that greeted me at the door is very friendly.

8. I met a lawyer. I consulted her about my legal issues. (Combine the sentences using ‘whom’)
I met a lawyer whom I consulted about my legal issues.

9. We hired a new secretary. I interviewed her last week. (Combine the sentences using ‘whom’)
We hired a new secretary whom I interviewed last week.

10. He introduced me to his sister. I have heard a lot about her. (Combine the sentences using ‘whom’)
He introduced me to his sister whom I have heard a lot about.

Exercise – 2

1. The CEO of the company is retiring. She has been with the company for thirty years. (Combine the sentences using ‘who’)
The CEO who has been with the company for thirty years is retiring.

2. The firefighter rescued the cat from the burning building. He received a bravery award. (Combine the sentences using ‘who’)
The firefighter who rescued the cat from the burning building received a bravery award.

3. I visited a museum. It displayed ancient artifacts. (Combine the sentences using ‘that’)
I visited a museum that displayed ancient artifacts.

4. The dress is made of silk. It is very expensive. (Combine the sentences using ‘that’)
The dress that is made of silk is very expensive.

5. The restaurant serves delicious sushi. It is always crowded on weekends. (Combine the sentences using ‘which’)
The restaurant, which is always crowded on weekends, serves delicious sushi.

6. The project is due next week. I haven’t started working on it yet. (Combine the sentences using ‘that’)
The project that I haven’t started working on yet is due next week.

7. The book is on the shelf. I borrowed it from the library. (Combine the sentences using ‘that’)
The book that I borrowed from the library is on the shelf.

8. The computer crashed during the presentation. I was using it at the time. (Combine the sentences using ‘that’)
The computer that I was using at the time crashed during the presentation.

9. This is a famous engineering college. I joined engineering in it. (Combine the sentences using ‘which’)
This is a famous engineering college in which I joined engineering.

10. I have a friend. I admire her dedication to her studies. (Combine the sentences using ‘whom’)
I have a friend whom I admire for her dedication to her studies.

Voice

Change the following sentences into passive voice.

Exercise – 1

1. The kids planted trees in the park.
Trees were planted in the park by the kids.

2. The teacher explained the lesson to the students.
The lesson was explained to the students by the teacher.

3. Mary wrote a letter to her grandmother.
A letter was written to her grandmother by Mary.

4. The chef prepared sushi for the customers.
Sushi was prepared for the customers by the chef.

5. The construction workers built a new skyscraper downtown.
A new skyscraper downtown was built by the construction workers.

6. The librarian organized the books on the shelves.
The books on the shelves were organized by the librarian.

7. James fixed the broken window in the living room.
The broken window was fixed by James in the living room.

8. The doctor examined the patient’s symptoms carefully.
The patient’s symptoms were examined carefully by the doctor.

9. The band playetf music at the school assembly.
Music was played at the school assembly by the band.

10. The janitor cleaned the hallways after school.
The hallways were cleaned after school by the janitor.

Exercise – 2

1. I cannot understand his attitude.
His attitude cannot be understood.

2. We nurture the children to build up a bright future.
The children are nurtured to build up a bright future.

3. I am managing the duties at the office as well as home. –
The duties are being managed at the office as will as home.

4. Let us recultivate the land.
Let the land be recultivated.

5. The government is providing the illiterate people with necessary educational amenities.
The illiterate people are being provided with necessary educational amenities by the government.

6. Have you ever used your common sense?
Has your common sense ever been used?

7. Our efforts will inspire the people for successful growth.
The people will be inspired by our efforts for successful growth.

8. Someone created the trouble.
The trouble was created.

9. Can you lift this sack of rice?
Can this sack of rice be lifted by you?

10. Every student has expressed willingness to join the group.
Willingness to join the group has been expressed by every student.

Exercise -1

1. He was exhausted. He couldn’t concentrate on his work. (Combine the sentences using ‘because’)
He couldn’t concentrate on his work because he was exhausted.

2. She was afraid. She didn’t want to go alone. (Combine the sentences using ‘as’)
She didn’t want to go alone as she was afraid.

3. The restaurant was crowded. We couldn’t find a table. (Combine the sentences using ‘because’)
We couldn’t find a table because the restaurant was crowded.

4. The box was very heavy. He could not lift it.(Combine the sentences using ‘so… that’)
The box was so heavy that he could not lift it.

5. He was very busy. He managed to finish the project. (Combine the sentences using ‘although’)
Although he was very busy, he managed to finish the project.

6. She was determined. She achieved her goal. (Combine the sentences using ‘because’)
She achieved her goal because she was determined.

7. The traffic was heavy. We arrived late at the party. (Combine the sentences using ‘when’)
We arrived late at the party when the traffic was heavy.

8. He was unwell. He still went to the meeting.(Combine the sentences using ‘although’)
Although he was unwell, he still went to the meeting.

9. The hike was challenging. They reached the summit. (Combine the sentences using ‘despite’)
They reached the summit despite the challenging hike.

10. She was excited. She couldn’t sleep all night. (Combine the sentences using ‘because’)
She couldn’t sleep all night because she was excited.

Exercise – 2

1. She was late. She missed the bus. (Combine the sentences using,‘because’)
She missed the bus because she was late.

2. He was feeling unwell. He went to see the doctor. (Combine the sentences using ‘as’)
He went to see the doctor as he was feeling unwell.

3. He was hungry. He decided to order pizza. (Combine the sentences using ‘because’)
He decided to order pizza because he was hungry.

4. The dog was hungry. It finished its food quickly. (Combine the sentences using ‘because’)
The dog finished its food quickly because it was hungry.

5. He was angry. He didn’t speak to anyone. (Combine the sentences using ‘as’)
As he was angry, he didn’t speak to anyone.

6. The water was cold. They went swimming anyway. (Combine the sentences using ‘although’)
They went swimming anyway although the water was cold.

7. She was exhausted. She couldn’t keep her eyes open. (Combine the sentences using ‘because’)
She couldn’t keep her eyes open because she was exhausted.

8. The movie was boring. They left halfway through. (Combine the sentences using ‘because’)
They left halfway through the movie because it was boring.

9. The weather was pleasant. They went for a picnic. (Combine the sentences using ‘because’)
They went for a picnic because the weather was pleasant.

10. He was excited. He couldn’t sit still. (Combine the sentences using ‘because’)
He couldn’t sit still because he was excited.

Suitable Prepositions

Fill in the blanks with suitable prepositions given in brackets.

Exercise – 1

1. The project was completed successfully _____ the team’s collaborative efforts. (in front of / because of / on behalf of)
2. _____ the new regulations, employees need to undergo additional training. (In spite of / According to / For the sake of)
3. She volunteered at the shelter _____ her love for animals. (by means of / due to / in addition to)
4. He attended the meeting _____ his boss who was out of town. (in front of / on behalf of / because of)
5. _____ the holiday season, the store extended its hours. (Because of / In addition to / According to)
6. The message was conveyed to the team _____ email. (by means of / in spite of / for the sake of)
7. She made the decision to move _____ her career prospects. (in front of / because of / for the sake of)
8. The event was cancelled _____ the lack of participants. (in front of / because of / according to)
9. He completed the assignment successfully _____ the detailed instructions. (in spite of / by means of / in addition to)
10. _____ the recent storm, the power outage lasted for hours. (Due to / On behalf of / According to) %
1) because of
2) According to
4) on behalf of
5) Because of
6) by means of
7) for the sake of
8) because of
9) by means of
10) Due to

Exercise – 2

1. The flowers are _____ the vase. (in / on / at)
2. She sat _____ the chair. (in / on / at)
3. The spider crawled _____ the ceiling. (in / on / at)
4. The book is _____ the shelf. (in / on / at)
5. The picture hangs _____ the wall. (in / on / at)
6. The keys are _____ the bag. (in / on / at)
7. We’ll meet you _____ the station. (in / on / at)
8. The hall rolled _____ the stairs. (in / on / at)
9. The plane flew _____ the clouds. (in / on / at)
10. The dog ran _____ the vard. (in / on / at)
1. in
2. on
3. on
4. in
5. on
6. in
7. at
8. on
9. in
10. in

Suitable Forms of the Verbs

Fill in the blanks in the following sentences using correct forms of the verbs given in brackets.

Exercise – 1

1. If I ______ (have) more time, I would travel around the world.
2. She would buy the house if she ______ (have) enough money.
3. If it ______ (rain) tomorrow, we will stay indoors.
4. I ______ (visit) my grandparents if I had a car.
5. If he ______ (study) harder, he would pass the exam.
6. We would go to the beach if it ______ (be) sunny.
7. If I ______ (be) you, I would take the job offer.
8. She would be happier if she ______ (live) closer to her family.
9. If they ______ (know) about the party, they would come.
10. If I ______ (be) taller, I could reach the top shelf.
3. rains
4. would visit
5. studied
6. were
7. were
8. lived
9. knew
10. were

Exercise – 2

1. Have you finished your work?
Yes, I ______ (finish) my work.
2. Wilson ______ (begin) travelling the world in 2008.
3. I ______ (know) her for nearly four years.
4. If I ______ (be) the Prime Minister, I would provide free education at all levels.
5. It’s time Vyshnavi ______ (go) to bed.
6. I ______ (know) her for nearly four years.
7. He ______ (pass) his examination in 2016.
8. I ______ (see) Raju this week.
9. I ______ (lose) my pencil yesterday.
10. I ______ (know) her for nearly four years.
1. have finished
2. began
3. have known
4. were
5. went
6. have known
7. passed
8. have seen
9. lost
10. have known

Giving a Suitable Advice or Suggestion

Exercise – 1

1. Your friend is going to visit Kerala. Advise him to learn a few words of Malayalam before he goes.
You had better learn a few words of Malayalam.

2. Your friend has eyesight problem. Advise him to see a doctor.
You should consult an ophthalmologist.

3. Your friend has lost his money in the beach. Advise him not to carry more money in open places.
You had better not to carry more money in open places.

4. Your friend has been wasting money on expensive clothes. Advise her to be more careful with her money.
It would be wise on your part not to waste your money on expensive clothes.

5. Your friend is feeling sick. Advise him to see a doctor.
It would be better for you to see a doctor.

6. Your friend appears to be a good runner. Advise him to join a sports’ school.
You should join a sports’ school.

7. Your friend who rides a motorcycle doesn’t use a helmet. Advise him to use a helmet.
It would be better for you to use a helmet.

8. A teacher has assigned a task in English to your friend which, he thinks, is very difficult. Advise him to try it.
You should try it.

9. Your friend is neglecting studies. Suggest him/her to work hard for better results.
You could wofk hard for better results.

10. Your cousin is very selfish. Suggest him/her to change his/her mindset.

Exercise – 2

You should take a walk every morning.

2. Your neighbour is watching TV at high volume. How do you ask him/her to reduce the volume in a polite way?
Would you kindly reduce the volume of your TV?

3. Your friend appears to be a good runner. Advise him to join a sports school.
You should join a sports school.

4. Your friend is not speaking in English. Advise him/her to speak in English.
You should speak in English.

5. Your brother who rides a motor-cycle does not use a helmet. Advise him to use a helmet.
It would be better for you to use a helmet.

6. You want your friend to post a letter for you. How would you ask in a polite way?
Would you mind posting this letter for me?

7. Your cousin is rather fat and wants to lose weight. Advise him to exercise every day.
You ought to exercise every day.

8. Your friend has been losing weight recently. Tell her that she should see a doctor.
You should/ought to see a doctor.

9. Your friend wants to settle in Bengaluru. Suggest him/her to learn the Kannada lan¬guage.
You could learn the Kannada language.

10. Of late your aunt is not feeling well. Suggest her to go for health check-up.
You could go for health check-up.

Changing a Sentence into a Polite Request

Change the following into a polite request.

Exercise – 1

1. You to your friend: “Lend me your car for the weekend.”
Would you mind lending me your car for the weekend?

2. You to your coworker: “Cover my shift tomorrow.”
Could you please cover my shift tomorrow?

3. You to your sibling: “Help me move this furniture.”

4. You to your neighbour: “Water my plants while I’m away.”
Would it be possible for you to water my plants while I’m away?

5. You to your classmate: “Share your notes from the last lecture.”
Would you mind sharing your notes from the last lecture?

6. You to your roommate: “Turn down the music a bit.”
Would you mind turning down the music a bit?

7. You to your partner: “Make dinner tonight.”
Could you please make dinner tonight?

8. You to your teacher: “Provide extra help with this topic.”
Would it be possible for you to provide extra help with this topic?

9. You to your boss: “Consider me for the upcoming project.”
Would you mind considering me for the upcoming project?

10. You to your parent: “Pick me up from the train station.”
Could you please pick me up from the train station?

Exercise – 2

Would you mind lending me your phone charger?

2. You to your coworker: “Pass me the stapler.”
Could you please pass me the stapler?

3. You to your sibling: “Help me with my laundry.”

4. You to your neighbour: “Keep it down a bit.”
Would you mind keeping it down a bit?

Would you mind sharing your textbook with me?

6. You to your roommate: “Ture off the lights when you leave.”
Would you mind turning off the lights when you leave?

7. You to your partner: “Pick up some milk on your way home.”
Would you mind picking up some milk on your way home?

8. You to your teacher: “Explain this concept one more time.”
Could you please explain this concept one more time?

9. You to your boss: “Provide me with some feedback on my report.”
Would it be possible for you to provide me with some feedback on my report?

10. You to your parents: “Drop me off at the library.”
Could you please drop me off at the library?

Identifying the Expression

What do the following sentences mean? Put a tick (✓) mark against the right answer.

Exercise -1

1. i) Would you mind lending me your pen for a moment?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission (✓)

ii) I’m really sorry for stepping on your foot.
A) Apologizing ( )
B) Ordering (✓)
C) Refusing permission ( )
D) Thanking ( )

2. i) I’m unable to understand this concept. Can you explain it to me?
A) Offering help! ( )
B) Refusing help ( )
C) Seeking permission ( )
D) Making a request (✓)

ii) Please pass me the remote control.
A) Apologizing ( )
B) Requesting (✓)
C) Complimenting ( )
D) Refusing permission ( )

3. i) Let me know if you need me to pick up anything from the store for you.
A) Offering help (✓)
B) Refusing help ( )
C) Seeking permission ( )

ii) I’m sorry, but I can’t help you go in without a ticket.
A) Apologizing ( )
B) Offering help ( )
C) Refusing permission (✓)
D) Ordering ( )

4. i) I appreciate the offer, but I think I’ve got it under control.
A) Offering help ( )
B) Refusing help (✓)
C) Seeking permission ( )

ii) Would you mind picking up some milk on your way home?
A) Apologizing ( )
B) Requesting (✓)
C) Ordering ( )
D) Refusing permission ( )

5. i) Could you please give me some pointers on how to improve my presentation?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission ( )

ii) I apologize if my comments offended you.
A) Apologizing (✓)
B) Offering help ( )
C) Ordering ( )
D) Thanking ( )

Exercise – 2

1. i) If you need any assistance with your project, feel free to ask.
A) Offering help (✓)
B) Refusing help ( )
C) Seeking permission ( )

ii) You must finish your chores before you can watch TV.
A) Apologizing ( )
B) Refusing permission ( )
C) Requesting ( )
D) Ordering (✓)

2. i) I’m not sure if I should go with option A or option B. What do you think?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission ( )

ii) I’m sorry, but I won’t be able to attend the meeting tomorrow.
A) Apologizing (✓)
B) Requesting ( )
C) Refusing permissoin ( )
D) Offering help ( )

3. i) If I find your dog untied again, I’ll report to the police.
A) Giving a warning (✓)
B) Making a suggestion ( )
C) Expressing surprise ( )
D) Making a statement ( )

ii) Could you please make sure to lock the door when you leave?
A) Apologizing ( )
B) Requesting (✓)
C) Refusing permission ( )
D) Offering help ( )

4. i) I don’t need any help right now, but thanks for offering.
A) Offering help ( )
B) Refusing help (✓)
C) Seeking permission ( )

ii) I’m really sorry, but I have to cancel our dinner plans tonight.
A) Apologizing ( )
B) Offering help ( )
C) Refusing permission ( )
D) Declining invitation (✓)

5. i) Is it okay if I take a break now?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission (✓)

ii) You need to finish your assignment before you can go out to play.
A) Apologizing ( )
B) Refusing permission ( )
C) Requesting ( )
D) Ordering (✓)

Synonyms

Read the paragraph and write the Synonyms of the underlined words choosing the words given in the box.

1. [sacred texts, obtain, monarchy, visited, hardships, eventually]
GAUTAMA Buddha (563 B.C. – 483 B.C.) began life as a prince named Siddhartha Gautama, in northern India. At twelve, he was sent away for schooling in the Hindu sacred scriptures (a) and four years later he returned home to marry a princess. They had a son and lived for ten years as befitted royalty (b). At about the age of twenty- five, the Prince, heretofore shielded from the sufferings (c) of the world, while out hunting chanced upon a sick man, then an aged man, then a funeral procession, and finally (d) a monk begging for alms.
a) sacred texts
b) monarchy
c) hardships
d) eventually

2. [mysterious, troubled, knowledge, conserved, mischievous, sacred]
These sights so moved him that he at once went out into the world to seek enlightenment (a) concerning the sorrows he had witnessed. He wandered for seven years and finally sat down under a peepal tree, where he vowed to stay until enlightenment came. Enlightened after seven days, he renamed the tree the Bodhi Tree (Tree of Wisdom) and began to teach and to share his new understandings. At that point he became known as the Buddha (the Awakened or the Enlightened). The Buddha preached his first sermon at the city of Benares, most holy (b) of the dipping places on the River Ganges; that sermon has been preserved (c) and is given here. It reflects the Buddha’s wisdom about one inscrutable (d) kind of suffering.
a) knowledge
b) sacred
c) conserved
d) mysterious

3. [visited, obtain, guarded, teacher, heal, complained]
Kisa Gotami repaired (a) to the Buddha and cried, “Lord and Master (b), give me the medicine that will cure (c) my boy.” The Buddha answered, “I want a handful of mustard- seed.” And when the girl in her joy promised to procure (d) it, the Buddha added, “The mustard- seed must be taken from a house where no one has lost a child, husband, parent or friend.”
a) visited
b) teacher
c) heal
d) obtain

4. [condiment, sympathized, insight, dear, recall, sorrow]
Poor Kisa Gotami now went from house to house, and the people pitied (a) her and said, “Here is mustard- seed; take it!” But when she asked, “Did a son or daughter, a father or mother, die in your family?” they answered her, “Alas! the living are few, but the dead are many. Do not remind (b) us of our deepest grief (c).” And there was no house but some beloved (d) one had died in it.
a) sympathized
b) recall
c) sorrow
d) dear

5. [tired, glimmered, compassion, roadside, ruled, mourning]
Kisa Gotami became weary (a) and hopeless, and sat down at the wayside (b) watching the lights of the city, as they flickered (c) up and were extinguished again. At last the darkness of the night reigned (d) everywhere.
a) tired
c) glimmered
d) ruled

6. [loneliness, similarity, refusal, eternal life, destiny, customary]
And she considered the fate (a) of men, that their lives flicker up and are extinguished again. And she thought to herself, “How selfish am 1 in my grief! Death is common (b) to all; yet in this valley of desolation (c) there is a path that leads him to immortality (d) who has surrendered all selfishness.”
a) destiny
b) customary
c) loneliness
d) eternal life

7. [quenched, gave up, cherished, narrow-minded, desperate, shimmered]
Kisa Gotami became weary and hopeless (a), and sat down at the wayside watching the lights of the city, as they flickered up and were extinguished (b) again. At last the darkness of the night reigned everywhere. And she considered the fate of men, that their lives flicker up and are extinguished again. And she thought to herself, “How selfish (c) am 1 in my grief! Death is common to all; yet in this valley of desolation there is a path that leads him to immortality who has surrendered (d) all selfishness.,”
a) despar ate
b) quenched
c) narrow-minded
d) gave up

8. [ceramic, devotion, humans, short, mature, isolation]
The Buddha said, “The life of mortals (a) in this world is troubled and brief (b) and combined with pain. For there is not any means by which those that have been born can avoid dying; after reaching old age there is death; of such a nature are living beings. As ripe (c) fruits are early in danger of falling, so mortals when born are always in danger of death. As all earthen (d) vessels made by the potter end in being broken, so is the life of mortals.
a) humans
b) short
c) mature
d) ceramic

9. [troubled, endangered, mourning, transient, exit, killing]
“Of those who, overcome by death, depart (a) from life, a father cannot save his son, nor kinsmen their relations. Mark! while relatives are looking on and lamenting (b) deeply, one by one mortals are carried off, like an ox that is led to the slaughter (c). So the world is afflicted (d) with death and decay, therefore the wise do not grieve, knowing the terms of the world.
a) exit
b) mourning
c) killing
d) troubled

10. [happy, crying, mourning, obstacle, sorrowing, opposite]
“Not from weeping (a) nor from grieving (b) will anyone obtain peace of mind; on the contrary (c), his pain will be the greater and his body will suffer. He will make himself sick and pale, yet the dead are not saved by his lamentation (d).
a) crying
b) sorrowing
c) opposite
d) mourning

Antonyms

Read the paragraph and match the words given in Column ‘A’ with the Antonyms in Column ‘B’.

1. GAUTAMA Buddha (563 B.C. – 483 B.C.) began life as a prince named Siddhartha Gautama, in northern India. At twelve, he was sent away for schooling in the Hindu sacred (a) scriptures and four years later he returned home to marry a princess. They had a son and lived for ten years as befitted (b) royalty. At about the age of twenty- five, the Prince, heretofore shielded (c) from the sufferings of the world, while out hunting chanced upon a sick (d) man, then an aged man, then a funeral procession, and finally a monk begging for alms.

 A B a) sacred 1) misfit b) befitted 2) healthy c) shielded 3) unholy d) sick 4) flickered 5) pious 6) exposed

a-3, b-1, c-6, d-2

2. These sights so moved him that he at once went out into the world to seek enlightenment concerning the sorrows he had witnessed. He wandered (a) for seven years and finally (b) sat down under a peepal tree, where he vowed to stay until enlightenment came. Enlightened (c) after seven days, he renamed the tree the Bodhi Tree (Tree of Wisdom) and began to teach and to share his new understandings. At that point he became known as the Buddha (the Awakened or the Enlightened). The Buddha preached his first sermon at the city of Benares, most holy of the dipping places on the River Ganges; that sermon has been preserved and is given here. It reflects the Buddha’s wisdom about one inscrutable kind of suffering (d).

 A B a) wandered 1) pleasure b) finally 2) acceptance c) enlightened 3) stayed d) suffering 4) initially 5) repaired 6) ignorant

a-3, b-4, c-6, d-1

3. Kisa Gotami had an only son, and he died. In her grief (a) she carried the dead (b) child to all her neighbours, asking them for medicine, and the people said, “She has lost (c) her senses. The hoy (d) is dead.”

 A B a) grief 1) joy b) dead 2) engaged c) lost 3) stayed d) boy 4) girl 5) found 6) alive

a-1, b-6, c-5, d-4

4. Kisa Gotami repaired to the Buddha and cried (a), “Lord and Master, give me the medicine that will cure my boy.” The Buddha answered (b), “I want a handful of mustard- seed.” And when the girl in her joy promised to procure (c) it, the Buddha added, “The mustard-seed must be taken from a house where no one has lost a child, husband, parent or friend.”(d)

 A B a) cried 1) questioned b) answered 2) laughed c) procure 3) sad d) friend 4) believe 5) enemy 6) surrender

a-2, b- 1, c-6, d-5

5. Poor Kisa Gotami now went from house to house, and the people pitied (a) her and said, “Here is mustard- seed; take it!” But when she asked, “Did a son or daughter, a father or mother, die (b) in your family?” they answered her, “Alas! the Hying are few, but the dead (c) are many. Do not remind us of our deepest (d) grief.” And there was no house but some beloved one had died in it.

 A B a) pitied 1) alive b) die 2) live c) dead 3) envied d) deepest 4) befriend 5) back 6) shallowed

a-3, b-2, c-1, d-6

6. Kisa Gotami became weary (a) and hopeless (b) and sat down at the wavside watching the lights of the city, as they flickered up and were extinguished again. At last the darkness (c) of the night reigned everywhere. And she considered the fate of men. that their lives flicker up and are extinguished again. And she thought to herself, “How selfish am I in my grief! Death is common to all; yet in this valley of desolation there is a path that leads him to immortalitv who has surrendered all selfishness.”(d)

 A B a) weary 1) selflessness b) hopeless 2) light c) darkness 3) hopeful d) selfishness 4) energetic 5) unhappy 6) incorrect

a-4, b-3, c-2, d-1

7. The Buddha said, “The life of mortals in this world is troubled (a) and brief (b) and combined with pain (c). For there is not any means by which those that have been born can avoid dying: after reaching old age there is death (d); of such a nature are
living beings.

 A B a) troubled 1) unreal b) brief 2) life c) pain 3) pleasure d) death 4) long 5) peaceful 6) unhealthy

a-5, b-4, c-3, d-2

8. As ripe (a) fruits are early in danger of falling, so mortals when born are always in danger (b) of death. As all earthen vessels made by the potter end in being broken, so is the life of mortals. Both young (c) and adult, both those who are fools and those who are wise (d). all fall into the power of death; all are subject to death.

 A B a) ripe 1) old b) danger 2) triumph c) young 3) safety d) wise 4) unripe 5) ignorant 6) foolish

a-4, b-3, c-1, d-6

9. “Of those who, overcome (a) by death, depart from life, a father cannot save (b) his son, nor kinsmen their relations. Mark! while relatives are looking (c) on and lamenting deeply, one by one mortals are carried off, like an ox that is led to the slaughter. So the world is afflicted with death and decay, therefore the wise do not grieve (d), knowing the terms of the world.

 A B a) overcome 1) anger b) save 2) unrelated c) looking 3) rejoice d) grieve 4) ignoring 5) endanger 6) triumph

a-6, b-5, c-4, d-3

10. “Not from weeping (a) nor from grieving (b) will anyone obtain peace of mind; on the contrary, his pain (c) will be the greater and his body will suffer. He will make himself sick and pale, yet the dead are not saved by his lamentation. He who seeks peace should draw out the arrow of lamentation, and complaint, and grief. He who has drawn out the arrow and has become composed will obtain peace of mind; he who has overcome all sorrow will become free from sorrow (d), and be blessed.”

 A B a) weeping 1) joy b) grieving 2) comfort c) pain 3) celebrating d) sorrow 4) smiling 5) disown 6) bad

a-4, b-3, c-2, d-1

Right Forms of the Words

Fill in the blanks in the following passage choosing the right form of the words given in brackets.

1. GAUTAMA Buddha (563 B.C. – 483 B.C.) began life as a prince named Siddhartha Gautama, in northern India. At twelve, he was sent away for schooling in the Hindu ______ (a) (sacred / sacredly / sacredness) scriptures and four years later he ______ (b) (return / returned / returning) home to ______ (c) (married / marriage / marry) a princess. They had a son and lived for ten years as befitted ______ (d) (royal / royally / royalty).
a) sacred
b) returned
c) marry
d) royalty

2. At about the age of twenty-five, the Prince, heretofore shielded from the ______ (a) (suffer / suffered / sufferings) of the world, while out ______ (b) (hunt / hunted / hunting) chanced upon a sick man, then an aged man, then a funeral procession, and ______ (c) (finale / finally / final) a monk begging for alms. These sights so moved him that he ;at once went out into the world to seek enlightenment concerning the sorrows he had ______ (d) (witness / witnessed / witnessing). He wandered for seven years and finally sat down under a peepal tree, where he vowed to stay until enlightenment came.
a) sufferings
b) hunting
c) finally
d) witnessed

3. Enlightened after seven days, he renamed the tree the Bodhi Tree (Tree of Wisdom) and began to teach and to share his new understandings. At that point he became ______ (a) (know / knew / known) as the Buddha (the Awakened or the Enlightened). The Buddha ______ (b) (preach / preached / preaching) his first sermon at the city of Benares, most ______ (c) (holy / holiest / holiness) of the dipping places on the River Ganges; that sermon has been preserved and is given here. It reflects the Buddha’s ______ (d) (wise / wisdom / wisely) about one inscrutable kind of suffering.
a) known
b) preached
c) holy
d) wisdom

4. At length, Kisa Gotami ______ (a) (meet / met / meeting) a man who ______ (b) (reply / replied / replying) to her request, “I cannot give thee medicine for thy child, but I ______ (c) (know / knew / known) a physician who can.” And the girl said, ” ______ (d) (Pray / Prayed / Praying) tell me, sir; who is it?” And the man replied, “Go to Sakyamuni, the Buddha.”
a) met
b) replied
c) know
d) Pray

5. Kisa Gotami ______ (a) (repair / repaired / repairing) to the Buddha and ______ (b) (cry / cried / crying), “Lord and Master, give me the medicine that will cure my boy.” The Buddha ______ (c) (answer / answered / answering), “I want a handful of mustard- seed.” And when the girl in her joy ______ (d) (promise / promised / promising) to procure it, the Buddha added, “The mustard-seed must be taken from a house where no one has lost a child, husband, parent or friend.”
a) repaired
b) cried
d) promised

6. Poor Kisa Gotami now ______ (a) (go / went / gone) from house to house, and the people pitied her and said, “Here is mustard-seed; take it!” But when she ______ (b) (ask / asked / asking), “Did a son or daughter, a father or mother, ______ (c) (die / died / dying) in your family?” they answered her, “Alas! the living are few, but the dead are many. Do not remind us of our ______ (d) (deep / deeper / deepest”) grief.” And there was no house but some beloved one had died in it.
a) went
c) die
d) deepest

7. Kisa Gotami became weary and ______ (a) (hopelessly / hopeless / hopelessness), and sat down at the wayside ______ (b) (watch / watched / watching) the lights of the city, as they ______ (c) (flicker / flickered / flickering) up and were extinguished again. At last the ______ (d) (darkness / dark / darkly) of the night reigned everywhere.
a) hopeless
b) watching
c) flickered
d) darkness

8. And she ______ (a) (consider / considered / considering) the fate of men, that their lives flicker up and are extinguished again. And she thought to herself, “How ______ (b) (selfishly / selfishness / selfish) am I in my grief! Death is common to all; yet in this valley of ______ (c) (desolation / desolate / desolated) there is a path that leads him to ______ (d) (mortality / immortality / immortals) who has surrendered all selfishness.”
a) considered
b) selfish
c) desolation
d) immortality

9. The Buddha said, “The life of mortals in this world is ______ (a) (trouble / troubled / troubling) and brief and ______ (b) (combine / combined / combining) with pain. For there is not any means by which those that have been born can ______ (c) (avoiding / avoided / avoid) dying; after reaching old age there is ______ (d) (death / dead / died); of such a nature are living beings.
a) troubled
b) combined
c) avoid
d) death

10. As ripe fruits are early in danger of ______ (a) (fall / falls / falling) , so mortals when born are always in danger of death. As all earthen vessels made by the potter end in being ______ (b) (break / broke / broken), so is the life of mortals. Both young and adult, both those who are fools and those who are ______ (c) (wise / wisely / wisdom), all fall into the ______ (d) (powerful / powerfully / power) of death; all are subject to death.
a) falling
b) broken
c) wise
d) power

11. “Of those who, overcome by death, ______ (a) (depart / departed / departing) from life, a father cannot save his son, nor kinsmen their relations. Mark! while relatives are ______ (b) (look / looked / looking) on and ______ (c) (lament / lamented / lamenting) deeply, one by one mortals are carried off, like an ox that is led to the slaughter. So the world is afflicted with death and decay, therefore the wise do not grieve, ______ (d) (know / knowing / known) the terms of the world.
a) depart
b) looking
c) lamenting
d) knowing

12. “Not from ______ (a) (weep / weeps / weeping) nor from ______ (b) (grieve / grieved / grieving) will anyone obtain peace of mind; on the contrary, his pain will be the greater and his body will ______ (c) (suffer / suffered / suffering). He will make himself sick and pale, yet the dead are not saved by his lamentation. He who seeks peace should draw out the arrow of lamentation, and complaint, and grief. He who has drawn out the arrow and has become ______ (d) (compose /composed / composing) will obtain peace of mind; he who has overcome all sorrow will become free from sorrow, and be blessed.”
a) weeping
b) grieving
c) suffer
d) composed

Vowel Clusters

Complete the spelling of the words choosing ‘ae’, ‘ee’, ‘ea’, ‘ei’, ‘oo’, ‘ou’, ‘ai’, ‘ia’, ‘ie’ io’, ‘oi’ ‘au’ ‘ue’ or ‘ui’.

Exercise – 1

1. At twelve, he was sent away for (a) sch_ _ling in the Hindu sacred scriptures and (b) f_ _r years later he returned home to marry a princess.
2. They had a son and lived for ten y_ _rs as befitted royalty.
3. At about the age of twenty-five, the Prince, heretofore (a) sh_ _lded from the sufferings of the world, while (b) t_ _hunting chanced upon a sick man, then an aged man, then a funeral procession, and finally a monk begging for alms.
4. These sights so moved him that he at once went out into the world to s_ _k enlightenment concerning the sorrows he had witnessed.
5. He wandered for seven years and finally sat down under a p_ _pal tree, where he vowed to stay until enlightenment came.
6. Enlightened after seven days, he renamed the (a) tr_ _the Bodhi Tree (Tree of Wisdom) and began to (b) t_ _ch and to share his new understandings.
7. The Buddha pr_ _ched his first sermon at the city of Benares, most holy of the dipping places on the River Ganges.
8. Kisa Gotami had an only son, and he d_ _d.
9. In her grief she carried the dead child to all her (a) n_ _ghbours, asking them for medicine, and the people said, “She has lost her senses. The boy is (b) d_ _d.”
10. At length, Kisa Gotami met a man who replied to her (a) req_ _st, “I cannot give thee medicine for thy child, but I know a (b) physic_ _n who can.”
1. a) schooling b)four
2. years
3. a) shielded b) out
4. seek
5. peepal
6. a) tree b) teach
7. preached
8. died
10. a) request b) physician

Exercise – 2

1. Kisa Gotami (a) rep_ _red to the Buddha and (b) cr_ _d, “Lord and Master, give me the medicine that will cure my boy.”
2. (a) P_ _r Kisa Gotami now went from house to house, and the people pit_ _d her.
3. “Here is mustard (a) s_ _d; take it!” But when she asked, “Did a son or (b) d_ _ghter, a father or mother, die in your family?” they answered her, “Alas! the living are few, but the dead are many.
4. Do not remind us of our (a) d_ _pest grief.” And there was no (b) h_ _se but some beloved one had died in it.
5. Kisa Gotami became (a) w_ _ry and hopeless, and sat down at the wayside watching the lights of the city, as they flickered up and were (b) exting shed again.
6. At last the darkness of the night r_ _gned everywhere.
7. And she considered the fate of men, that their lives flicker up and are (a) exting_ _shed (b) ag_ _n.
8. And she (a) th_ _ght to herself, “How selfish am I in my (b) gr_ _f! Death is common to all; yet in this valley of desolation there is a path that leads him to immortality who has surrendered all selfishness.”
9. The Buddha said, “The life of mortals in this world is (a) tr_ _bled and (b) br_ _f and combined with pain.
10. As all (a) rthen_ _vessels made by the potter end in (b) b ng broken, so is the life of mortals.
11. Both (a) y_ _ng and adult, both those who are (b) f_ _ls and those who are wise, all fall into the power of death; all are subject to death.
12. Mark! while relatives are looking on and lamenting (a) d_ _ply, one by one mortals are carried off, like an ox that is led to the (b) si_ _ghter.
1. a) repaired b) cried
2. a) Poor b) pitied
3. a) seed b) daughter
4. a) deepest b) house
5. a) weary b) extinguished
6. reigned
7. a) extinguished b) again
8. a) thought b) grief
9. a) troubled b) brief
10. a) earthen b) being
11. a) young b) fools
12. a) deeply b) slaughter

Suffixes or Inflections

Fill in the blanks with correct suffixes given in the brackets.

1. So the world is afflicted with death and decay, therefore the wise do not grieve, know___(n / ing) the terms of the world.
2. “Not from weeping nor from griev___(ing / ance) will anyone obtain peace of mind; on the contrary, his pain will be the greater and his body will suffer.
3. He will make himself sick and pale, yet the dead are not saved by his lamentat___(ed / ion).
4. He who seeks peace should draw out the arrow of lamentat___(ed / ion), and complaint, and grief.
5. He who has drawn out the arrow and has become compos___(ing / ed) will obtain peace of mind; he who has overcome all sorrow will become free from sorrow, and be blessed.”
6. He returned home to marry a princ___(es / ess)
7. He chanced upon a sick man, then an aged man, then a funeral proce___. (sion / ssion)
8. It reflects the Buddha’s wisdom about the inscrut___(able / ible) kind of suffering.
9. She carried the dead child to all her neighb___. (urs / ours)
10. I cannot give th^ medi___(sine /cine) (a) for the child, but I know a physi___. (sian /cian) (b)
11. I want a hand___(ful / full) of mustard-seed.
12. The dead are not saved by his lamenta___. (tion / sion)
1. knowing
2. grieving
3. lamentation
4. lamentation
5. composed
6. princess
7. procession
8. inscrutable
9. neighbours
10. (a) medicine (b) physician
11. handful
12. lamentation

Identifying the Wrongly Spelt Word

Find the wrongly spelt word and write the correct spelling.

Exercise – 1

1. afflicted
2. sufferings
3. procession
4. enlightenment
5. extinguished
6. surrendered
7. grieving
8. significant
9. prevent
10. survivors

Exercise – 2

1. religious
2. fatigue
3. awakened
4. dictionary
5. medicine
6. preached
7. befitted
8. daughter
9. complaint
10. flickered

Dictionary Skills

1. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘depart’?
verb

b) What is the opposite of ‘depart’?
arrive

2. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘enlighten’?
verb

b) What is the adjective form of ‘enlighten’?
enlightening

3. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘extinguish’?
verb

b) What is the synonym of ‘extinguish’?
put out

4. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘holy’?

b) What is the opposite of ‘holy’?
unholy

5. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the synonym of the word ‘hopeless’?
terrible

b) What is the adverb form of ‘hopeless’?
hopelessly

6. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘immortal’?

b) What is the opposite of ‘immortal’?
mortal

7. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘inscrutable’?

b) What is the adverb form of ‘inscrutable’?
inscrutably

8. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘joyful’?

b) What is the noun form of ‘joyful’?
joyfulness

9. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘lament’?
verb & noun

b) What are the synonyms of the word ‘lament’?
lament, bemoan

10. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the opposite of ‘mortal’?
immortal

b) What is the synonym of ‘mortal’?

11. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the opposite of ‘selfish’?
unselfish, selfless

b) What is the noun form of ‘selfish’?
selfishness

12. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the synonym of the word ‘sorrow’?
grief

b) What is the adjective form of the word ‘sorrow’?
sorrowful

13. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘weary’?

b) What is the noun form of ‘weary’?
weariness

Classification of Words

Arrange the following words under correct hedings.

1. [seeds, alms, world, city, vessel, fruit, house, neighbourhood]

 Things Places 1. 1. 2. 2. 3. 3. 4. 4.

 Things Places 1. seeds 1. world 2. alms 2. city 3. vessel 3. house 4. fruit 4. neighbourhood

2. [son, child, daughter, father, mother, adult, young, aged man]

 Relations Phases in the life of a man 1. 1. 2. 2. 3. 3. 4. 4.

 Relations Phases in the life of a man 1. daughter 1. child 2. son 2. young 3. father 3. adult 4. mother 4. aged man

3. [potter, physician, a sick man, an aged man, master, prince, a monk, a funeral procession]

 People The four sights Siddhartha had seen 1. 1. 2. 2. 3. 3. 4. 4.

 People The four sights Siddhartha had seen 1. potter 1. a sick man 2. physician 2. an aged man 3. prince 3. a funeral procession 4. master 4. a monk

4. contentment, fatigue, thirst, heat, love, pain, anger, disgust

 Emotions Physical sensations 1. 1. 2. 2. 3. 3. 4. 4.

 Emotions Physical sensations 1. contentment 1. pain 2. love 2. heat 3. anger 3. thirst 4. disgust 4. fatigue

Matching

Match the words / phrases given under Column ‘A’ with their meanings given under Column ‘B’.
1.

 Column – A Column – B 1) chance upon A) affected by suffering, disease or pain 2) at length B) to become mad 3) afflicted with C) damaged with 4) to lose one’s senses D) come across by chance E) after a long time F) at lengthwise

1-D, 2-E, 3-A, 4-B

2.

 Column – A Column – B 1) sermon A) a speech 2) valley of desolation B) religious or moral talk 3) mortals C) a state ot high spiritual knowledge 4) enlightenment D) those bound to die E) those that have no death F) an area which is filled with deep sorrow

1-B, 2-F, 3-D, 4-C

3.

 Column – A Column – B 1) be composed A) went to 2) lamentation B) to obtain something especially with difficulty 3) procure C) affected by suffering, disease or pain 4) inscrutable D) made or formed from several parts, things or people E) expression of sorrow F) something which cannot be understood

1-D, 2-E, 3-B, 4-F

Conversation

Question 1.
You have read the lesson ‘The Sermon at the Benares’.
Now write a possible conversation between you and your father regarding the journey of Siddhartha Gautama from prince to Buddha.
Son : Hey Dad, 1 have just read a fascinating lesson called “The Sermon at Benares” about Siddhartha Gautama’s journey from prince to Buddha. Have you heard about it?

Father : Yes, I’m familiar with the story of Siddhartha Gautama. It’s quite remarkable, isn’t it? What specifically caught your interest?

Son : Well, I found it incredible how Siddhartha, despite being sheltered from the world’s suffering, was deeply moved by the sight of sickness, old age, and death. It prompted him to seek enlightenment. Can you imagine having everything and then choosing to leave it all behind to find answers to life’s deepest questions?

Father : Absolutely, it takes immense courage and introspection to embark on such a profound spiritual journey. Siddhartha’s realization that material wealth and worldly pleasures couldn’t shield him from the realities of human suffering is a powerful lesson for us all.

Son : I was also struck by the moment when Siddhartha sat under the Bodhi Tree and vowed to stay until enlightenment came. It’s inspiring how he remained steadfast in his determination despite facing numerous challenges and temptations along the way.

Father : Yes, his perseverance and unwavering commitment to seeking truth are admirable qualities. It’s a reminder that the path to enlightenment is often fraught with obstacles, but it’s the inner strength and resilience that ultimately lead to spiritual awakening.

Son : And the part about Kisa Gotami’s story, seeking medicine for her dead child, only to realize the universality of suffering and death, was quite poignant. It shows that suffering is a shared human experience, and ; true liberation comes from transcending selfishness and embracing compassion for all beings.

Father : Indeed, the story of Kisa Gotami beautifully illustrates the Buddha’s teachings on the nature of suffering and the impermanence of life. It emphasizes the importance of empathy and understanding in our interactions with others, especially during times of grief and loss.

Son : Reading about Siddhartha Gautama’s transformation into the Buddha has given me a lot to think about. It’s a profound reminder of the power of self-discovery, compassion, and the quest for spiritual enlightenment.

Father : I’m glad you found it thought-provoking, son. Siddhartha Gautama’s journey is a timeless tale that continues to inspire and resonate with people from all walks of life. It’s a testament to the enduring relevance of his teachings in today’s world.

Son : Absolutely, Dad. Thanks for sharing your insights. It’s always enlightening to discuss these profound philosophical concepts with you.

Father : Anytime, son. I’m always here to explore these ideas with you. It’s what keeps our minds and spirits enriched.

Question 2.
You have read the poem ‘For Anne Gregory’. Consider the societal pressures, particularly those faced by women, to conform to certain beauty standards. How do these standards affect individuals’ perceptions of themselves and their relationships? What is your opinion about it?
Write a possible conversation between you and your mother.
I : Mom, I’ve been thinking a lot about societal pressures, especially those that women face to conform to certain beauty standards: It seems like these standards can really affect how women perceive themselves and their relationships.

Mother : That’s a very relevant topic, dear. Society often imposes unrealistic expectations on women, especially regarding their appearance. From a young age, girls are bombarded with images and messages about what it means to be beautiful.

I : Exactly, and it’s not just about physical appearance – it’s about hair colour, body shape, skin tone, and so much more. Women are constantly comparing themselves to these unrealistic standards, which can lead to feelings of inadequacy and low self-esteem.

Mother : Absolutely. It’s heartbreaking to see how these standards can impact women’s self-worth and confidence. They may feel like they’re not good enough unless they fit into a narrow definition of beauty, which is completely unfair and unrealistic.

I : And it’s not just about how women see themselves – it also affects their relationships. If someone feels like they have to change their appearance to be loved or accepted, it can create a lot of insecurity ; and instability in their relationships.

Mother : That’s true. It’s sad to think that some people may only value women for their external appearance rather than their inner qualities and character. True love and acceptance should be based on who a person is, not what they look like.

I : Absolutely, Mom. I think it’s important for us to challenge these societal norms and redefine what beauty means. We need to celebrate diversity and encourage women to embrace their unique qualities rather than trying to conform to impossible standards.

Mother : I couldn’t agree more, dear. As women, we should empower each other to love and accept ourselves just as we are, regardless of what society tells us. It’s time to break free from these limiting beliefs and embrace our true beauty, inside and out.

Question 3.
You have understood Bholi’s courage and self-assertion in rejecting societal norms
and standing up for herself, despite facing ridicule and discrimination.
Now write a possible conversation between you and your teacher regarding this.
I : Good morning, Madam. 1 wanted to talk to you about the story we read, “Bholi”. I found her character incredibly inspiring, especially in the way she stood up for herself despite all the challenges she faced.

Teacher : Good morning! I’m glad to hear that you found Bholi’s character inspiring. Yes, her courage and self-assertion are indeed remarkable considering the circumstances she was in. What specifically stood out to you about her actions?

I : Well, I was amazed by how Bholi refused to conform to societal expectations, especially regarding her marriage. Even though she was pressured and ridiculed, she remained true to herself and spoke up against injustice.

Teacher : Absolutely. Bholi’s refusal to marry someone she didn’t respect or love, simply to satisfy societal norms, shows tremendous strength of character. It’s not easy to go against the grain, especially in a society where conformity is highly valued.

I : Exactly. And what struck me the most was how Bholi found her voice despite being labeled as unintelligent and unattractive. She defied those stereotypes and proved that inner strength and resilience matter more than outward appearances.

Teacher : That’s a profound observation. Bholi’s journey highlights the importance of self-worth and inner resilience. Despite the ridicule and discrimination she faced, she never lost faith in herself or her abilities. It’s a powerful reminder that our worth isn’t determined by how others perceive us.

I : Absolutely. Bholi’s story made me reflect on the importance of standing up for what we believe in, even when it’s difficult. It’s a lesson i I’ll carry with me, especially when facing challenges or discrimination in my own life.

Teacher : I’m glad to hear that Bholi’s story resonated with you on such a deep level. It’s stories like hers that remind us of the power of resilience, courage, and self-assertion in the face of adversity. Thank you for sharing your thoughts on this remarkable character.

I : Thank you, Madam, for guiding us through this story and helping us uncover its deeper meanings. It’s been a truly eye-opening experience.

Teacher : You’re very welcome. It’s discussions like these that make teaching so rewarding. If you ever have more thoughts or questions about “Bholi” or any other stories we study, feel free to reach out.

Diary Entry

Question 1.
Kisa Gotami, who had lost her son had been enlightened by the Buddha. After attaining the spiritual knowledge grief-stricken Kisa Gotami realised that there would be no use lamenting over the dead ones. She is now unattached and wise enough to draw out the lamentation and grief from her heart. She attained peace now. She wanted to write the diary entry of her feelings.
Imagine yourself to be Kisa Gotami and write your feelings as diary entry.
20 May, 533 B.C.
Friday, 9.30 p.m.

Dear Diary,
How peaceful I am now! The flames of my grief have been extinguished by the realisation of spiritual wisdom. The waves of turbulent sea of misery have been suppressed by the enlightenment of soul. Losing a young son is heartbreaking and melancholic. I suffered a lot of agony, dolefulness and sombre. I felt emptiness of life after I lost my son. Life is desolated, dreary and purposeless. But Sakyamuni has brought my soul back. His preachings have meant a lot for me. If he had not been there, I would have still been wading through sinful, mystic, and mythic sea of mortal life.

He has enlightened my heart. With his preachings I came to know that death is common to all. No one can avoid death. No one can be saved from death. Not from weeping, nor from grieving will anyone attain peace of mind. Moreover by lamenting over the dead one we will suffer a lot of pain and will be weak. We cannot attain peace of mind from grieving and weeping. With our lamentation the dead can not be saved. One who has drawn out the sorrow from his mind only can become free from sorrow. Thanks to Sakyamuni for enlightening my soul, for bringing peace to my mind.
Kisa Gotami

Question 2.
Bholi’s parents were dumbfounded for her courage, confidence and prowess. They were made speechless by her eloquent and fluent outburst. They were astonished how their stammering, nervous and timid simpleton Bholi has grown to be a girl of wisdom, self-reliance, and independence. Especially her mother was astounded at her astonishing transformation. She wanted to write her feelings as a diary entry.
Imagine yourself to be Bholi’s mother and write the diary entry. Assume her mother’s name is Lakshmi.
25 November, 20xx
Saturday, 9.30 p.m.

Dear Diary,
I am astonished, astounded and dumbfounded at our Bholi’s eloquent, fluent and unstammered outburst. I never expected that she can talk that. She is now grown up to be confident, wise, self-reliant and independent, I think it is all because of education she has got in that school. We must be thankful to the teachers and the headmistress. ’ They have transformed our stammering, witless stupid simpleton into well-learned, well-mannered and empowered. The only thing we have done good for Bholi was to join her in the school. I can never be excused for I have not treated Bholi with love and affection. I have been unconcerned, indifferent towards her. I did not treat her as I had treated other children. I thought she was witless and ugly with pock-marks all over face and body. I have never bought new clothes for her. I used to give her worn out clothes of Champa. Her father made me realise our mistake. He told me that our carelessness, apathy and our unconcerned attitude were the causes of her fate.

How intelligent she has grown up now! She has unmasked Bishamber’s greedy, contemptible and deceptive attitude. I am not downcast now for her marriage with Bishamber Nath ended in failure. Moreover I am very much happy because my girl now is self-reliant and intelligent. We now look for a capable young man who likes her for her education, abilities and skills but not for looks. I am now proud of my girl. From now onwards I keep loving her. Instead of showing hatred towards us she wanted to take care of us in our old age. How concerned was she towards us ! I would like to apologise to her. I wanted to show our affection towards her. Bholi, – no – no -1 don’t call her Bholi – she is our Sulekha. Sulekha, excuse me.
Lakshmi

Letter Writing

Question 1.
Books in libraries are essential for education and knowledge dissemination. Libraries offer access to a vast array of books, providing opportunities for intellectual growth and personal enrichment. Through books, readers can explore new ideas, cultures, and perspectives, fostering empathy and understanding. Moreover, books preserve cultural heritage and collective wisdom, ensuring that future generations have access to the accumulated knowledge and insights of humanity.
Write a letter to your school librarian recommending new books for the library.
[City, State, ZIP Code]
[Phone Number]
[Date]
[Librarian’s Name]
[School Library Name]
[City, State, ZIP Code]
Dear [Librarian’s Name],
I hope this letter finds you well. As a dedicated advocate for the importance of reading and education, I am writing to recommend some new books for our school library, specifically targeted towards 10th standard students.

Books play a pivotal role in shaping the intellectual curiosity and academic growth of students, offering them opportunities to explore diverse topics, expand their knowledge, and foster a love for reading. With that in mind, I believe that the following titles would be valuable additions to our library’s collection:

1. “To Kill a Mockingbird” by Harper Lee
• This timeless classic offers profound insights into themes of justice, morality, and compassion, making it an essential read for students exploring complex social issues.

2. “The Catcher in the Rye” by J.D. Salinger
• Salinger’s iconic novel provides a poignant portrayal of adolescence and identity, resonating with students as they navigate the challenges of growing up and finding their place in the world.

3. “The Diary of a Young Girl” by Anne Frank
• Anne Frank’s diary offers a poignant and powerful firsthand account of life during the Holocaust, serving as a testament to the resilience of the human spirit and the importance of hope in the face of adversity.

4. “1984” by George Orwell
• Orwell’s dystopian masterpiece explores themes of surveillance, government control, and individual freedom, prompting critical reflection on the nature of power and authority in society.

5. “The Alchemist” by Paulo Coelho
• Coelho’s allegorical tale of self-discovery and personal transformation inspires readers to pursue their dreams and embrace the journey of life with courage and resiliency.

I believe that these books not only align with the academic curriculum but also offer valuable insights into the human experience, encouraging students to engage critically with literature and develop empathy and understanding for diverse perspectives.

I am more than willing to assist in the acquisition and organization of these books in the library and look forward to seeing them become accessible to our students. Thank you for considering my recommendations, and I am eager to contribute to the enrichment of our school library’s collection.
Warm regards,

Question 2.
Public transportation plays a crucial role in society by providing accessible and sustainable mobility options for communities. It reduces traffic congestion, air pollution, and carbon emissions, contributing to a cleaner environment and healthier cities. Public transit enhances economic opportunities by connecting people to jobs, education, and essential services. Additionally, public transportation reduces the dependency on personal vehicles, saving time and money for commuters.
Pen a letter to your local councillor advocating for better public transportation.
[City, State, ZIP Code]
[Phone Number]
[Date]
[Local Councilor’s Name]
[City, State, ZIP Code]

Dear [Councilor’s Name],
I hope this letter finds you in good health and spirits. As a concerned member of our community, 1 am writing to advocate for the improvement and expansion of public transportation services in our area.

Public transportation plays a vital role in promoting accessibility, sustainability, and economic prosperity within our communities. By offering affordable and convenient mobility options, public transit reduces traffic congestion, mitigates air pollution, and lowers carbon emissions, contributing to a cleaner environment and healthier cities for current and future generations.

Moreover, public transportation serves as a lifeline for many individuals and families, connecting them to essential services, employment opportunities, educational institutions, and recreational facilities. Accessible and reliable public transit is particularly crucial for those who may not have access to private vehicles or alternative means of transportation, ensuring equal opportunities for all members of our community to participate in social, economic, and cultural activities.

However, despite its numerous benefits, our current public transportation system faces several challenges, including limited coverage, insufficient frequency, and inadequate infrastructure. As a result, many residents continue to experience difficulties in accessing transportation services, leading to increased reliance on personal vehicles and exacerbating traffic congestion and environmental degradation.

By investing in the improvement and expansion of public transportation, we can create a more inclusive, sustainable, and vibrant community where all residents have access to safe, affordable, and reliable mobility options. I urge you to consider these recommendations and take proactive steps towards building a better future for our community.

Thank you for your attention to this important issue, and I look forward to your support and leadership in advocating for better public transportation in our area. Warm regards,

Question 3.
Offering condolences for the loss of a loved one is a gesture of comfort and support during a challenging time. Expressing heartfelt sympathy acknowledges the pain and grief of those mourning. Sharing kind words, memories, or simply being present can provide solace and reassurance. It offers hope and reassurance during the healing process.
Write a letter to a Mend offering condolences for the loss of a loved one.
[City, State, ZIP Code]
[Phone Number]
[Date]
[Friend’s Name]
[City, State, ZIP Code]

Dear [Friend’s Name],
I hope this letter finds you surrounded by love and support during this incredibly difficult time. I was deeply saddened to hear about the loss of your [relationship of the deceased], [Name of the deceased]. Please accept my heartfelt condolences, and know that my thoughts and prayers are with you and your family as you navigate through this period of grief and sorrow.

Losing someone dear to us is undoubtedly one of life’s most profound and challenging experiences. The pain of your loss is immeasurable, and no words can fully ease the heaviness of your heart. However, please know that you are not alone in your sorrow. You have a circle of friends and loved ones who care deeply about you and are here to offer you comfort and support in any way we can.

I want to takte a moment to honor the memory of [Name of the deceased] and the beautiful legacy they leave behind. [He/She] was a remarkable [relationship of the deceased], and [his/her] presence brought warmth, kindness, and joy to all who knew [him/her]. The memories and moments shared with [him/her] will forever remain cherished treasures in our hearts.

During this time of mourning, please remember to be gentle with yourself and allow yourself to feel whatever emotions come naturally. Grieving is a deeply personal journey, and there is no right or wrong way to navigate through it. Lean on the love and support of your family and friends, and allow yourself the time and space you need to heal.

If there is anything at all that I can do to support you during this time, please do not hesitate to reach out. Whether it’s a listening ear, a shoulder to lean on, or simply a comforting presence, know that I am here for you, now and always.

May you find peace and solace in the loving memories you shared with [Name of the deceased], and may [his/her] spirit continue to shine brightly in your heart, guiding you through the darkness towards the light of healing and hope.
With deepest sympathy and love,

Biographical Sketch

Question 1.
Write a biographical sketch of Dr. Sreedhara Somanath, the Chairman of ISRO using the following information.
Name : Dr. Sreedhara Somanath
Full Name : Sreedhara Panicker Somanath
Date of Birth : 7 July, 1963
Place of Birth : Alappuzha (Alleppey), Kerala
Parents : Father : Vedamparambil Sreedhara Panicker, a Hindi teacher Mother : Thankamma
Spouse : Valsala Kumari; Children : 2
Education : High School: St. Augustine’s High School, Aroor
Mechanical Engineering : TKM College of Engineering, Kollam Master’s Degree in Aerospace Engineering from the Indian Institute of Science
Ph.D in Mechanical Engineering from the Indian Institute of Technology, Madras
Profession : Joined ISRO in 1985
Positions help : Served as the Project Manager of PSLV Project
Served as the Project Director for GSLV Mk – III
Under his leadership ISRO successfully launched the Chandrayaan – 3 Mission in 2023
The spacecraft launched on July 14, 2023 successfully fulfilled its mission on 23 August, 2023 at 6.04 p.m.
Awards : Space Gold Medal from the Astronautical Society of India (ASI);
Performance Excellence Award and Team Excellence Award from ISRO for his contributions to GSLV Mk – III
Dr. Sreedhara Somanath is the Chairman of ISRO. He is a prominent Indian aerospace engineer and rocket technologist. His full name is Sreedhara Panicker Somanath. He was born on 7th July, 1963. He was born at Alappuzha (Alleppey), Kerala. His father was Vedamparambil Sreedhara Panicker, a Hindi teacher. His mother was Thankamma. His wife’s name is Valsala Kumari. He had two children. He had his high school education at St. Augustine’s High School, Aroor. He completed his Mechanical Engineering in TKM College of Engineering, Kollam. He took Master’s Degree in Aerospace Engineering from the Indian Institute of Science.

He did his Ph.D in Mechanical Engineering from the Indian Institute of Technology, Madras. He joined ISRO 1985. He served as the Project Manager of PSLV Project. He served as the Project Director for GSLV Mk – III. Under his leadership ISRO successfully launched Chandrayan – 3 Mission in 2023. The spacecraft was launched on July 14, 2023 and successfully completed its mission on 23 August, 2023 at 6.04 p.m. He received Space Gold Medal from the Astronautical Society of India (ASI). He received Performance Excellence Award and Team Excellence Award from ISRO for his contributions to GSLV Mk – III.

Question 2.
Write a biographical sketch of Chandrabose – the famous Telugu lyricist using the following information.
Name : Chandrabose
Full Name : Kanukuntla Subhash Chandrabose
Born : On 10 May 1978 at Challagariga, Warangal District
Parents : Father : Narasaiah – a primary school teacher
Spouse : Suchitra
Children : 2; Son : Nanda Vanamali; Daughter : Amrutha
Education : B.E in Electrical and Electronics Engineering from J[NTU, Hyderabad
Debut film : Taj Mahal in 1995
Total songs written : 3600 for 850 films in a span of 25 years
Prominence : for writing lyrics of the song ‘Naatu Naatu’ for the film RRR that won the Academy Award – 2023 and Golden Globe Award for Best Original Song – 2022
Awards & Honours : 2023 Best Original Song Award for ‘Naatu Naatu’ from RRR Critics Choice Award for ‘Naatu Naatu’ in 2022
Nandi Awards : 2 (in 2002 and 2004)
Filmfare Award for Best Lyricist in 2014 and 2018 SIIMA Awards in 2014, 2018 and 2022
Chandrabose is the famous Telugu lyricist. His full name is Kanukuntla Subhash Chandrabose. He was born on 10 May, 1978. He was born at Challagariga in Warangal District. His father is Narasaiah – a primary school teacher. His mother is Madhunamma. His spouse is Suchitra. He has two children. His son is Nanda Vanamali and his daughter is Amrutha. He studied B.E. in Electrical and Electronics – Engineering in JNTU, Hyderabad. His debut film was Taj Mahal in 1995. He has written 3600 songs for 850 films in a span of 25 years.

He became world famous for writing lyrics of the song ‘Naatu Naatu’ for the film RRR. It has won the Academy (Oscar) Award in the best original song category for the year 2023. His song ‘Naatu Naatu’ has won Golden Globe Award for the best original song in 2023. He received Critics Choice Award for ‘Naatu Naatu’ song in 2022. He won Nandi Awards for 2 times in 2002 and 2004. He won Filmfare Award for the best lyricist in 2014 and 2018. He won SIIMA Awards in 2014, 2018 and 2022 as the best lyricist.

Question 3.
Write a biographical sketch of S.S. Rajamouli, the famous Indian film director using the following information.
Name : S.S. Rajamouli; other names : Jakkanna; Raja; Nandi
Full Name : Koduri Srisaila Sri Rajamouli
Birth : On October 10, 1973; at Amareshwara Camp in Raichur District, Karnataka
Parents : Father : K.V. Vijayendra Prasad – a screen writer
Mother : Raja Nandini – a housewife
Education : Primary education : in Kowur,
Higher education: Eluru
Engineering : C.R. Reddy College, Eluru
Spouse : Rama Rajamouli
Children : 2; S.S. Karthikeya and S.S. Mayookha
Profession : Film director
Career : Started as assistant to K. Raghavendra Rao
directed soap operas; directed TV series ‘Santhi Nivasam’
Debut : Student No. 1 as a director
Awards : Best feature film for Eega in 2012
Best feature film for Bahubali – The Beginning in 2015
Best popular film Bahubali – 2 in 2017
Filmfare Awards : Best director for 4 times
A.P’s Nandi Awards : For the best direction for 3 times
International Awards & Accolades : The song ‘Naatu Naatu’ for the film RRR has won Academy (Oscar) Award and Golden Globe Award
Listed among 100 most influential people of 2023 by the Time Magazine
Best Director for Bahubali – The Beginning
Best Director at the New York Film Critics Circle Award for RRR in 2023
Honours : Padma Shri in 2016
S.S. Rajamouli is the famous Indian film director. His full name is Koduri Srisaila Sri Rajamouli. His other names are Jakkanna, Raja and Nandi. He was born on October 10, 1973. He was born at Amareswara Camp in Raichur District, Karnataka. His father is K.V. Vijayendra Prasad – a screen writer. His mother’s name is Raja Nandini – a housewife. He had his primary education at Kowur. He had his higher education at Eluru. He completed his engineering from C.R. Reddy College, Eluru. His spouse is Rama Rajamouli. He has two children. His son is SS Karthikeya and his daughter is SS Mayookha. He is a professional film director. He started his career as an assistant to the famous director K. Raghavendra Rao. He directed soap operas. He directed TV Series ‘Santhi Nivasam’.

His debut film as a director was Student No. 1. His famous films were Vikramarkudu, Eega, Yamadonga, Magadheera, Maryada Ramanna, Bahubali 1 & 2 and RRR. His film ‘Eega’ was chosen as the best feature film in 2012. His film ‘Bahubali – The Beginning’ won as the best feature film in 2015. His Bahubali – 2 was named as the best popular film for 2017. He won the Filmfare Best Director Award for 4 times. He won A.P’s Nandi Award for the Best Direction for 3 times. The song ‘Naatu Naatu’ for the film RRR has won Academy (Oscar) Award for the film RRR. This song has also won the Golden Globe Award. He was listed among 100 most influential people of 2023 by the Time Magazine. He won the Best Director Award for Bahubali – The Beginning. He won the Best Director Award at the New York Film Critics Circle Award for RRR in 2023. He was honoured with Padma Shri in 2016 by the Government of India.

Framing ‘Wh’ Questions

Read the following passage carefully focussing on the underlined parts.

1. Her name was Sulekha (A), but since her childhood everyone had been calling her Bholi (B), the simpleton (C). She was the fourth daughter of Numberdar Ramlal (D). When she was ten months old, she had fallen off the cot on her head (E) and perhaps it had damaged some part of her brain.
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) What was her name?
B) What had she been called since her childhood?
C) What do you mean by ‘Bholi’?
D) What was her father’s name?
E) What had happened to her when she was ten months old?

2. At birth, the child was very fair and pretty (A). But when she was two years old she had an attack of small-pox (B). Only the eyes were saved, but the entire body was permanently disfigured by deep black pock-marks (C). Little Sulekha (D) could not speak (E) till she was five.
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) How was the child when she was born?
B) What happened to her when she was two years old?
C) What happened to her because of small-pox?
D) Who could not speak till she was five?
E) What would Sulekha not do?

3. But Ramlal was worried about Bholi (A). She had neither good looks nor intelligence (B).
Bholi was seven years old (C) when Mangala was married. The same year a primary school for girls (D) was opened in their village. The Tehsildar sahib (E) came to perform its opening ceremony.
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) About whom was Ramlal worried?
B) Why was Ramlal worried about Bholi?
C) How old was Bholi when Mangala was married?
D) What was opened in their village?
E) Who came to perform its opening ceremony?

4. When they reached the school, the children were already in their classroom (A). Ramlal handed over his daughter to the headmistress (B). Left alone, the poor girl looked about her with fear-laden eyes (C). There were several rooms, and in each room girls like her squatted on mats, reading from books or writing on slates (D). The headmistress asked Bholi to sit down in a corner (E) in one of the classrooms.
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) Where were the children when they reached the school?
B) Whom did Ramlal hand over his daughter?
C) How did the girl look about her?
D) What were the girls doing?

5. Bishambar Nath was a well-to-do grocer (A). He came with a big party of friends and relations with him (B) for the wedding (C). Ramlal was overjoyed to see such pomp and splendour (D). He had never dreamt that his fourth daughter would have such a grand wedding (E).
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) What was Ramlal?
B) What did he come with for wedding?
C) What did he come for?
D) What was Ramlal overjoyed to see?
E) What had he never dreamt?

6. On Bishamber’s greedy face appeared a triumphant smile (A). He had gambled and won. “Give me the garland,” he announced. Once again the veil (B) was slipped back from the bride’s face, but this time her eyes were not downcast (C). She was not looking up, looking straight at her prospective husband (D), and in her eyes there was neither anger nor hate, only cold contempt (E).
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) What appeared on Bishamber’s greedy face?
B) What was slipped back from the bride’s face?
C) How were her eyes this time?
D) What was she looking straight at?
E) What was there in her eyes?

7. Bishamber raised the garland to place it round the bride’s neck (A); but before he could do so, Bholi’s hand struck out like a streak of lightning (B) and the garland was flung into the fire. She got up and threw away the veil (C).

“Pitaji !” said Bholi in a clear loud voice (D); and her father, mother, sisters and brothers, relations and neighbours were startled to hear her speak without even the slightest stammer (E).
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) Why did Bishamber raise the garland?
B) How did Bholi’s hand strike?
C) What did she do?
D) How did Bholi speak?
E) What were they startled to speak?

8. Ramlal stood rooted to the ground, his head bowed low with the weight of grief and shame (A). The flames of the sacred fire slowly died down. Everyone was gone. Ramlal turned to Bholi and said, “But what about you, no one will ever marry you now. What shall we do with you?” (B)

And Surekha said in a voice that was calm and steady (C), “Don’t you worry, Pitaji! In your old age I will serve you and Mother (D) and I will teach in the same school where I learnt so much (E).”
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) How did his head bow low with?
B) What did Ramlal say to Bholi?
C) How was Sulekha’s voice?
D) What will Sulekha want to do?
E) Where will Sulekha want to teach?

Information Transfer

Question 1.
Study the following bar graph carefully and write a paragraph based on the information given in it.

The bar graph illustrates the steady growth in the number of employees within a company over a five-year period, showcasing a positive trajectory in workforce expansion. Starting from 100 employees in 2017, the company experienced consistent annual increases, reaching 120 employees in 2018, 150 employees in 2019, 180 employees in 2020, and peaking at 200 employees in 2021. This upward trend indicates the company’s resilience and capacity for expansion, reflecting its ability to attract and retain talent while meeting evolving business demands. The progressive increase in workforce size suggests a flourishing organizational environment, potentially driven by factors such as business growth, market demand, and strategic workforce planning. Overall, the data underscores the company’s commitment to sustained growth and development, positioning it favorably for future opportunities and challenges.

Question 2.
Study the following pie chart carefully and write a paragraph based on the information given in it.

The pie chart offers insights into the distribution of pet preferences among individuals, highlighting the varying degrees of popularity among different types of pets. Dogs emerge as the most favored choice, representing 40% of pet ownership, underscoring their status as beloved companions in many households. Following closely behind are cats, constituting 30% of pet preferences, reflecting their enduring appeal and widespread popularity as domesticated animals. Fish claim a significant but somewhat smaller portion at 15%, indicative of their popularity among hobbyists and enthusiasts. Birds and reptiles secure smaller slices of the pie at 10% and 5% respectively, suggesting a niche but devoted following among pet owners who appreciate the unique characteristics and care requirements of these animals. The pie chart illuminates the diverse array of preferences within the realm of pet ownership, reflecting the multifaceted nature of human-animal relationships and the varied roles that pets play in people’s lives.

Question 3.
The chart below gives the percentage of social media users by age in India in 2020. Write a paragraph analysing the given information.

The given table shows the internet activities done by certain age groups. The first age group in teens who do not do any product research. Instead, the teens indulge themselves in other activities more like online games, searching for news, doing online shopping etc. An average teen tends to look for entertainment on the internet more than something knowledgeable. Then come the people in their 20s. Their internet activity is drastically different from the teens, and they can be seen being indulged in more product research, news, shopping, searching for people etc. It can be seen through the given statistics that a person’s mentality towards internet changes as they grow, and they tend to go for productive things to do there. The people in their 30s have a similar activity status to the 20s people except for a decrease in downloads, searching for people and online games.

The given data reflects that the usage of internet is the need of the hour. It is not only bound to the age group be it a 10-year-old or an above seventy. Internet is a part and parcel of everyone’s life.

## AP 10th Class English 7th Lesson Important Questions

These AP 10th Class English Important Questions 7th Lesson will help students prepare well for the exams.

## AP Board 10th Class English 7th Lesson Important Questions and Answers

Study Skills

I. Study the following table carefully.

 Year Petrol Consumption (Million Litres) 2018 45600 2019 48200 2020 42500 2021 50300 2022 46800

1) In which year did India experience the highest petrol consumption?
2021

2) What was the petrol consumption in 2020?
42500 million litres

3) How much petrol was consumed in 2021? (C)
A) 48,200 million litres
B) 42,500 million litres
C) 50,300 million litres
C) 50,300 million litres

4) In which year was there a decrease in petrol consumption compared to the previous year? (C)
A) 2018
B) 2019
C) 2020
C) 2020

5) What is the difference in petrol consumption between 2019 and 2022? (A)
A) 1,400 million litres
B) 2,500 million litres
C) 3,600 million litres
A) 1,400 million litres

II. Read the Pie chart carefully.

1) What percentage of time do the students spend on socializing during their weekends?
25%

2) How much time do the students spend on chores and exercise combined?
15% + 10% = 25%

3) Which activity consumes the least amount of time during the weekends? (A)
A) Leisure
B) Exercise
C) Socializing
A) Leisure

4) If the students spend 10 hours on study during the weekend, how many hours do they spend on socializing? (B)
A) 6.25 hours
B) 8.75 hours
C) 10 hours
B) 8.75 hours

5) What percentage of time is allocated to study and socializing combined? (C)
A) 45%
B) 50%
C) 65%
C) 65%

III. Read the bar chart carefully.

1) How many people prefer autumn as their favourite season?
30 people

2) What is the total number of people surveyed regarding their favourite seasons?
150 people (40 + 55 + 30 + 25)

3) Which season is the most favoured among the people surveyed? (B)
A) Spring
B) Summer
C) Autumn
B) Summer

4) How many more people prefer summer than winter? (B)
A) 20
B) 30
C) 40
B) 30

5) How many people prefer either spring or autumn? (C)
A) 60
B) 65
C) 70
C) 70

IV. Study the following table carefully.

1) Who had the fastest time in the 100m sprint?
Charlie

2) What was David’s long jump distance?
6.1 m

3) Who had the longest shot put distance? (B)
A) Alice
B) Charlie
C) David
B) Charlie

4) What was the average shot put distance among all athletes? (B)
A) 12.3m
B) 12.4m
C) 12.5m
B) 12.4m

5) Which athlete had the best overall performance? (C)
A) Alice
B) Bob
C) Charlie
C) Charlie

V. Read the Pie chart carefully.

1) What percentage of students prefer Soccer?
35%

2) How many students prefer Swimming?
10%

3) What is the percentage of students who prefer Basketball? (A)
A) 25%
B) 20%
C) 30%
A) 25%

4) If there are 200 students in total, how many prefer Tennis? (A)
A) 40
B) 50
C) 30
A) 40

5) What proportion of students prefer Volleyball? (A)
A) 10%
B) 15%
C) 20%
A) 10%

Section – B

Grammar & Vocabulary

Combining Sentences Using who/which/that, etc.

Exercise – 1

1. The girl won the science fair. She built a solar-powered car. (Combine the sentences using ‘who’)
The girl who bujlt a solar-powered car won the science fair.

2. My friend is a talented musician. She plays the piano beautifully. (Combine the sentences using‘who’)
My friend, who plays the piano beautifully, is a talented musician.

3. The doctor treated my grandfather. She has been practicing medicine for twenty years. (Combine the sentences using ‘who’)
The doctor who has been practicing medicine for twenty years treated my grandfather.

4. The movie received critical acclaim. It had an unexpected plot twist. (Combine the sentences using ‘which’)

5. The book is a bestseller. It explores the mysteries of time travel. (Combine the sentences using ‘that’)
The book that explores the mysteries of time travel is a bestseller.

6. She made a promise. She would always be there for her friends. (Combine the sentences using ‘that’)
She made a promise that she would always be there for her friends.

7. The tree gives good shade. It is a tamarind tree. (Combine the sentences using ‘that’)
The tree that gives good shade is a tamarind tree.

8. I met a new colleague. I admire her dedication to her work. (Combine the sentences using ‘whom’)
I met a new colleague whom I admire for her dedication to her work.

9. I have a neighbour. I often borrow tools from him. (Combine the sentences using ‘whom’)
I have a neighbour from whom I often borrow tools.

10. We have a tour guide. I highly recommend her for your trip. (Combine the sentences using ‘whom’)
We have a tour guide whom I highly recommend for your trip.

Exercise – 2

1. Sarah is an excellent photographer. She captured stunning landscapes during her recent trip. (Combine the sentences using ‘who’)
Sarah, who captured stunning landscapes during her recent trip, is an excellent photographer.

2. The chef prepared our delicious dinner. She has won multiple culinary awards. (Combine the sentences using ‘who’)
The chef who has won multiple culinary awards prepared our delicious dinner.

3. I bought a new laptop. It has a powerful graphics card. (Combine the sentences using ‘that’)
I bought a new laptop that has a powerful graphics card.

4. The company launched a new product. It incorporates cutting-edge technology. (Combine the sentences using ‘that’)
The company launched a new product that incorporates cutting-edge technology.

5. The car broke down. It was just serviced last week. (Combine the sentences using ‘that’)
The car that was just serviced last week broke down.

6. The restaurant is famous for its pasta. I tried it last night. (Combine the sentences using ‘that’)
The restaurant that I tried last night is fcimous for its pasta.

7. The dog barked loudly. It woke up the entire neighbourhood. (Combine the sentences using ‘that’)
The dog that barked loudly woke up the entire neighbourhood.

8. I have a friend. He can speak six languages fluently. (Combine the sentences using ‘who’)
I have a friend who can speak six languages fluently.

9. The student submitted an impressive essay. I assigned him a difficult topic. (Combine the sentences using ‘whom’)
The student to whom I assigned a difficult topic submitted an impressive essay.

10. She introduced me to a famous author. I have always wanted to meet him. (Combine the sentences using‘whom’)
She introduced me to a famous author whom I have always wanted to meet.

Voice

Change the following sentences into passive voice.

Exercise – 1

1. The dog chased the cat up the tree.
The cat was chased up the tree by the dog.

2. Sarah baked cookies for the school fundraiser.
Cookies were baked for the school fundraiser by Sarah.

3. The company launched a new product last month.
A new product was launched by the company last month.

4. The teacher assigned homework over the weekend.
Homework was assigned over the weekend by the teacher.

5. John fixed the leaky faucet in the kitchen.
The leaky faucet in the kitchen was fixed by John.

6. The kids built a sandcastle at the beach.
A sandcastle was built at the beach by the kids.

7. The gardener planted flowers in the backyard.
Flowers were planted in the backyard by the gardener.

8. The chef cooked a delicious meal for the guests.
A delicious meal was cooked for the guests by the chef.

9. The team won the championship trophy.
The championship trophy was won by the team.

10. The mechanii repaired the car’s engine.
The car’s engine was repaired by the mechanic.

Exercise – 2

1. The boy has picked up weight.
Weight has been picked up by the boy.

2. The thief scared the land owner with his knife.
The land owner was scared by the thief with his knife.

3. Are you fetching some water for me?
Is some water being fetched by you for me?

4. I saw him alighting the bus.
He was seen alighting the bus.

5. The doctor was examining the inpatients.
The inpatients were being examined by the doctor.

6. The old man has locked up the lifelong earning^ in a safe place.
The lifelong earnings have been locked up in a safe place by the old man.

7. I cannot bear this swindling.
This swindling cannot be borne by me.

8. Pay the fees within two days.
The fees is to be paid within two days.

9. You are riding a wild horse.
A wild horse is being ridden by you.

10. The host has received all the invitees cordially.
All the invitees have been received cordially by the host.

Exercise – 1

1. She is diligent. She gets good grades. (Combine the sentences using ‘as’)
She gets goodjgrades as she is diligent.

2. It was raining heavily. We decided to stay indoors. (Combine the sentences using ‘because’)
We decided to stay indoors because it was raining heavily.

3. He studied hard. He passed the exam. (Combine the sentences using ‘since’)
He passed the exam since he studied hard.

4. The book was very interesting. I finished it in one sitting. (Combine the sentences using ‘so…that’)
The book was so interesting that I finished it in one sitting.

5. She was tired. She continued working on her project. (Combine the sentences using ‘although’)
Although she was tired, she continued working on her project.

6. The traffic was terrible. He arrived late for the meeting. (Combine the sentences using ‘when’)
He arrived late for the meeting when the traffic was terrible.

7. He forgot his umbrella. He got soaked in the rain. (Combine the sentences using ‘because’)
He got soaked in the rain because he forgot his umbrella.

8. She practised the piano every day. She became a skilled pianist. (Combine the sentences using ‘because’)
She became a skilled pianist because she practised the piano every day.

9. He was very ill. He went to work anyway. (Combine the sentences using ‘although’)
Although he was very ill, he went to work anyway.

10. The movie was scary. I watched it until the end. (Combine the sentences using ‘although’)
Although the movie was scary, I watched it until the end.

Exercise – 2

1. The weather was cold. They decided to stay indoors. (Combine the sentences using ‘because’)
They decided to stay indoors because the weather was cold.

2. She was hungry. She ate a sandwich. (Combine the sentences using ‘because’)
She ate a sandwich because she was hungry.

3. He missed the bus. He arrived late for school. (Combine the sentences using ‘when’)
He arrived late for school when he missed the bus.

4. We are planning to see a film. It’s a holiday tomorrow. (Combine the sentences using ‘since’)
Since it is a holiday tomorrow, we are planning to see a film.

5. The road was flooded. We had to take a detour. (Combine the sentences using ‘because?)

6. The music was loud. They couldn’t hear each other. (Combine the sentences using ‘because’)
They couldn’t hear each other because the music was loud.

7. She was short of money. She couldn’t buy the dress. (Combine the sentences using ‘because’)
She couldn’t buy the dress because she was short of money.

8. He was tired. He couldn’t finish his homework. (Combine the sentences using ‘as’)
He couldn’t finish his homework as he was tired.

9. The cat was hungry. It meowed loudly. (Combine the sentences using ‘as’)
The cat meowed loudly as it was hungry.

10. The baby was asleep. We tiptoed around the room. (Combine the sentences using ‘while’)
We tiptoed around the room while the baby was asleep.

Suitable Prepositions

Fill in the blanks with suitable prepositions given in brackets.

Exercise – 1

1. She was praised ______ her dedication to the project. (in front of / by means of / for the sake of)
2. ______ the heavy rain, the match continued. (Because of / According to / On behalf of)
3. He decided to resign his dissatisfaction with the management. (in addition to / due to / in front of)
4. The teacher spoke ______ the class about the upcoming exam. (in front of / according to / on behalf of)
5. The book was written ______ the author’s research on the topic. (in spite of / because of / by means of)
6. She attended the meeting ______ her colleagues.(in addition to / due to / on behalf of)
7. ______ the instructions, the project was completed successfully. . (Because of / According to / In spite of)
8. He accepted the award ______ his team’s efforts. (for the sake of / in front of / because of)
9. She went on stage ______ her fear of public speaking. (in spite of / due to / by means of)
10. They travelled to Italy ______ the beautiful scenery. (in front of / for the sake of / according to)
1) by means of
2) Because of
3) due to
4) in front of
5) by means of
6) on behalf of
7) According to
8) because of
9) in spite of
10) for the sake of

Exercise – 2

1. The plane flies ______ the clouds. (In / on / at)
2. We had a picnic ______ the park. (in / on/ at)
3. The phone is ______ the table. (In / on / at)
4. The fish swim ______ the water. (in / on / at)
5. They live ______ a big house. (In / on / at)
6. The keys are ______ the door. (In / on / at)
7. The ball is ______ the basket. (In / on / at)
8. We’ll meet you ______ the restaurant. (In / on / at)
9. The cat is sleeping ______ the couch. (in / on / at)
10. The car is parked ______ the garage. (in / on / at)
1. in
2. in
3. on
4. in
5. in
6. at
7. in
8. at
9. on
10. in

Suitable Forms of the Verbs

Fill in the blanks in the following sentences using correct forms of the verbs given in brackets.

Exercise – 1

1. The cake ______ (bake) by my mother.
2. English ______ (speak) all over the world.
3. The letter ______ (send) yesterday.
4. The window ______ (break) by the strong wind.
5. The house ______ (clean) every week.
6. The book ______ (write) by J.K. Rowling.
7. The problem ______ (solve) by the team.
8. The car ______ (repair) by the mechanic.
9. The new bridge ______ (construct) by the government now.
10. The test ______ (take) by the students next Monday.
1. is baked
2. is spoken
3. was sent
4. was broken
5. is cleaned
6. was written
7. was solved
8. was repaired
9. is being constructed
10. will be taken

Exercise – 2

1. I ______ (throw) away most of my old books when I ______ (move) to new house.
2. I ______ (know) him most of my working life.
3. We ______ (belong) to the tennis club in the village we used to live in.
4. Chinese craftsmen ______ (invent) both paper and printing.
5. Tom ______ (be) ill since Christmas.
6. She ______ (open) the door and (sit) down quickly.
7. She ______ (sleep) very badly whenever she stayed alone.
8. We ______ (meet) each other when I ______ (work) in a restaurant.
9. He ______ (take) out the cake, before the cooker ______ (whistle).
10. I ______ (be) tired because I ______ (walk) fast.
1. threw, moved
2. have known
3. belonged
4. invented
5. has been
6. opened, sat

Giving a Suitable Advice or Suggestion

Exercise – 1

1. Your sibling keeps procrastinating on their homework. Advise him/her to manage their time better.
It’s important to manage your time effectively, especially when it comes to homework.

2. Your colleague is frequently late for work. Advise him/her to set alarms and leave home earlier.
Setting alarms and leaving home earlier can help you arrive at work on time consistently.

3. Your neighbour is struggling to lose weight. Advise him/her to focus on a balanced diet and regular exercise routine.
Focusing on a balanced diet and incorporating a regular exercise routine can help you achieve your weight loss goals.

4. Your classmate is stressed about upcoming exams. Advise him/her to create a study schedule and take breaks to relax.
Creating a study schedule and taking regular breaks to relax can help alleviate stress during exam preparation.

5. Your friend is overspending and constantly running out of money. Advise him/her to create a budget and stick to it.

Prioritizing tasks and knowing when to ask for help can alleviate feelings of overwhelm with work and personal commitments.

7. Your teammate is struggling with a difficult project. Advise him/her to break it down into smaller tasks and seek assistance if necessary.
Breaking down the project into smaller tasks and seeking assistance when needed can make it more manageable and less overwhelming.

8. Your sibling is having trouble making friends in a new school. Advise him/her to join clubs or extracurricular activities where they share interests with others.
Joining clubs or extracurricular activities where you share interests with others can help you make friends in a new school.

9. Your roommate is having trouble sleeping at night. Advise him/her to establish a bedtime routine and create a comfortable sleep environment.
Establishing a bedtime routine and creating a comfortable sleep environment can help improve your quality of sleep at night.

10. Your colleague is having difficulty expressing their ideas during team meetings. Advise him/her to practice public speaking and participate in group discussions more actively.
Practicing public speaking and actively participating in group discussions can help improve your ability to express ideas during team meetings.

Exercise – 2

1. Your friend is constantly forgetting important appointments. Advise him/her to use a calendar or scheduling app to stay organized.
Using a calendar or scheduling app can help you remember important appointments more effectively.

2. Your classmate is struggling to stay motivated to study. Advise him/her to set specific goals and reward themselves for achieving milestones.
Setting specific goals and rewarding yourself for achieving milestones can help maintain motivation to study.

Starting with simple recipes and gradually trying more complex dishes can help improve your cooking skills over time.

4. Your coworker is dealing with a difficult boss. Advise him/her to communicate openly and professionally about their concerns.

5. Your neighbour is struggling to quit smoking. Advise him/her to seek support from friends, family, or a smoking cessation programme.
Seeking support from friends, family, or a smoking cessation programme can help you quit smoking successfully.

6. Your teammate wants to improve their productivity at work. Advise him/her to prioritize tasks and minimize distractions during work hours.
Prioritizing tasks and minimizing distractions during work hours can help improve productivity in the workplace.

7. Your cousin is unsure about which career path to pursue. Advise him/her to explore different options through internships, volunteering, or informational interviews.
Exploring different options through internships, volunteering, or informational interviews can help you decide on a career path.

8. Your roommate is feeling homesick after moving to a new city. Advise him/her to stay connected with friends and family through regular calls or visits.
Staying connected with friends and family through regular calls or visits can help alleviate feelings of homesickness.

9. Your colleague is struggling to keep up with technology advancements in their field. Advise him/her to enroll in online courses or attend workshops to update their skills.

10. Your friend wants to start a healthier lifestyle but doesn’t know where to begin. Advise him/her to start with small changes such as eating more fruits and vegetables and incorporating physical activity into their daily routine.
Starting with small changes such as eating more fruits and vegetables and incorporating physical activity into your daily routine can help you start a healthier lifestyle.

Changing a Sentence into a Polite Request

Change the following sentence into a polite request.

Exercise – 1

1. You to your colleague: “Hand me the file.”
Could you please hand me the file?

2. You to your sibling: “Pass me the remote.”
Would you mind passing me the remote?

3. You to your waiter: “Bring me some water.”
Could you kindly bring me some water?

Would it be possible for you to lend me your notes?

Could you please return my book?

6. You to your teammate: “Give me a hand with this project.”
Would you mind giving me a hand with this project?

7. You to your partner: “Pick up some groceries on your way home.”
Could you kindly pick up some groceries on your way home?

Would it be possible for you to extend the deadline for the assignment?

9. You to your friend: “Drop me off at the station.”
Could you please drop me off at the station?

10. You to your boss: “Grant me permission to take a day off.”
Would it be possible for you to grant me permission to take a day off?

Exercise – 2

Would you mind lending me your pen?

3. You to your sibling: “Could you help me with my homework?”
Would you mind helping me with my homework?

4. You to your colleague: “Could you cover for me during the meeting?”
Would it be possible for you to cover for me during the meeting?

5. You to your parents: “Could you drive me to the airport?”
Could you please drive me to the airport?

6. You to your teammate: “Could you review my presentation?”
Would you mind reviewing my presentation?

7. You to your landlord: “Could you fix the leaky faucet?”
Could you please fix the leaky faucet?

8. You to your partner: “Could you pick up dinner on your way home?”
Would you mind picking up dinner on your way home?

9. You to your teacher: “Could you clarify this concept for me?”
Could you please clarify this concept for me?

10. You to your boss: “Could you provide feedback on my project proposal?”
Would it be possible for you to provide feedback on my project proposal?

Identifying the Expression

What do the following sentences mean? Put a tick (✓) mark against the right answer.

Exercise – 1

1. i) Can we leave now?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission (✓)

ii) I am sorry, I can’t let you go in without a gate pass.
A) Apologizing ( )
B) Refusing permission (✓)
C) Ordering ( )
D) Refusing help ( )

2. i) Do you need any assistance with that?
A) Offering help (✓)
B) Refusing help ( )
C) Seeking permission ( )

ii) Could you please pass me the salt?
A) Apologizing ( )
B) Requesting (✓)
C) Complimenting ( )
D) Ordering ( )

3. i) I’m not sure what to do next. Any suggestions?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission ( )

ii) I apologize for arriving late to the meeting.
A) Apologizing (✓)
B) Thanking ( )
C) Ordering ( )
D) Refusing permission ( )

4. i) Would you like me to help you carry those bags?
A) Offering help (✓)
B) Refusing help ( )
C) Seeking permission ( )

ii) You must finish your homework before going out to play.
A) Apologizing ( )
B) Refusing permission ( )
C) Ordering (✓)
D) Requesting help ( )

5. i) No, thank you. I can handle it on my own.
A) Offering help ( )
B) Refusing help (✓)
C) Seeking permission ( )

ii) You must attend the classes.
A) Offering ( )
B) Suggestion ( )
C) Obgligation (✓)
D) Giving information ( )

Exercise – 2

1. i) May I borrow your umbrella?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission (✓)

ii) Please close the door behind you when you leave.
A) Apologizing ( )
B) Requesting (✓)
C) Ordering ( )
D) Offering help ( )

2. i) Could you please give me some guidance on this project?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission ( )

ii) Would you mind turning down the music a bit?
A) Apologizing ( )
B) Requesting (✓)
C) Ordering ( )
D) Refusing permission ( )

3. i) I don’t think I need your assistance right now.
A) Offering help ( )
B) Refusing help (✓)
C) Seeking permission ( )

ii) I’m really sorry, but I won’t be able to attend the party.
A) Apologizing (✓)
B) Requesting ( )
C) Refusing permission ( )
D) Ordering ( )

4. i) Do you have any recommendations for a good restaurant?
A) Offering help ( )
B) Refusing help ( )
C) Seeking permission ( )

ii) You have to wear a helmet while riding your bike.
A) Apologizing ( )
B) Refusing permission ( )
C) Requesting ( )
D) Ordering (✓)

5. i) Let me know if you need a hand with anything.
A) Offering help (✓)
B) Refusing help ( )
C) Seeking permission ( )

ii) I’m sorry, I cannot accept your invitation this time.
A) Apologizing ( )
B) Refusing permission (✓)
C) Ordering ( )
D) Declining invitation ( )

Synonyms

Read the paragraph and write the Synonyms of the underlined words choosing the words given in the box.

1. [hobby, entrance, acquired, Mends, shining, detailed]
There was a girl named Valliammai who was called Valli for short. She was eight years old and very curious about things. Her favourite pastime (a) was standing in the front doorway (b) of her house, watching what was happening in the street outside. There were no playmates (c) of her own age on her street, and this was about all she had to do.
But for Valli, standing at the front door was every bit as enjoyable as any of the elaborate (d) games other children played.
a) hobby
b) entrance
c) friends
d) detailed

2. [sneaked, season, blocked, ignite, longingly, colloquial]
Day after day she watched the bus, and gradually a tiny wish crept (a) into her head and grew there: she wanted to ride on that bus, even if just once. This wish became stronger and stronger, until it was an overwhelming desire. Valli would stare wistfully (b) at the people who got on or off the bus when it stopped at the street corner. Their faces would kindle (c) in her longings, dreams, and hopes. If one of her friends happened to ride the bus and tried to describe the sights of the town to her, Valli would be too jealous to listen and would shout, in English: “Proud! proud!” Neither she nor her friends really understood the meaning of the word, but they used it often as a slang (d) expression of disapproval.
a) sneaked
b) longingly
c) ignite
d) colloquial

3. [ceiling, cautious, acquired, protruding, barely, wealth]
Over many days and months Valli listened carefully to conversations between her neighbours and people who regularly used the bus, and she also asked a few discreet (a) questions here and there. This way she picked up (b) various small details about the bus journey. The town was six miles from her village. The fare was thirty paise one way – “which is almost nothing at all,” she heard one well-dressed man say, but to Valli, who scarcely (c) saw that much money from one month to the next, it seemed a fortune (d).
a) cautious
b) acquired
c) barely
d) wealth

4. [gazed, departing, upper, authoritatively, extending, move slowly]
Well, one fine spring day the afternoon bus was just on the point of leaving (a) the village and turning into the main highway when a small voice was heard shouting: “Stop the bus! Stop the bus!” And a tiny hand was raised commandingly (b).
The bus slowed down to a crawl (c), and the conductor, sticking (d) his head out the door, said,
a) departing
b) authoritatively
c) move slowly
d) extending

5. [notice, windscreen, ceiling, shining, irritated, glowed]
It was a new bus, its outside painted a gleaming (a) white with some green stripes along the sides. Inside, the overhead (b) bars shone (c) like silver. Directly in front of Valli, above the windshield (d), there was a beautiful clock. The seats were soft and luxurious.
a) shining
b) ceiling
c) glowed
d) windscreen

6. [gazed, trench, gnawing, blocked, consumed, fabric]
Valli devoured (a) everything with her eyes. But when she started to look outside, she found her view cut off (b) by a canvas (c) blind that covered the lower part of her window. So she stood up on the seat and peered (d) over the blind.
a) consumed
b) blocked
c) fabric
d) gazed

7. [waterway, village, meadow, anxious, far off, trench]
The bus was now going along the bank of a canal (a). The road was very narrow. On one side there was the canal and, beyond it, palm trees, grassland (b), distant (c) mountains, and the blue, blue sky. On the other side was a deep ditch (d) and then acres and acres of green fields – green, green, green, as far as the eye could see.
a) waterway
c) far-off
d) trench

8. [settlement, surprised, creep, notice, irritated, worried]
Suddenly she was startled (a) by a voice. “Listen, child,” said the voice, “you shouldn’t stand like that. Sit down.”
Sitting down, she looked to see who had spoken. It was an elderly man who had honestly been concerned (b) for her, but she was annoyed (c) by his attention (d).
a) surprised
b) worried
c) irritated
d) notice

9. [disgusting, auricles, suppressed, sprinted, areca nut, gnawing]
Valli found the woman absolutely repulsive (a) – such big holes she had in her ear lobes (b), and such ugly earrings in them! And she could smell the betel nut (c) the woman was chewing (d) and see the betel juice that was threatening to spill over her lips at any moment.
a) disgusting
b) auricles
c) areca nut
d) gnawing

10. [thorough, unwanted, economically, suppressed, huge, clang]
Her first journey – what careful, painstaking (a), elaborate plans she had had to make for it! She had thriftily (b) saved whatever stray (c) coins came her way, resisting every temptation to buy peppermints, toys, balloons, and the like, and finally she had saved a total of sixty paise. How difficult it had been, particularly that day at the village fair, but she had resolutely stifled (d) a strong desire to ride the merry- go-round, even though she had the money.
a) thorough
b) economically
c) unwanted
d) suppressed

11. [settlement, swallowing, barriers, walker, hooted, enormous]
The bus rolled on now cutting across a bare landscape, now rushing through a tiny hamlet (a) or past an odd wayside shop. Sometimes the bus seemed on the point of gobbling (b) up another vehicle that was coming towards them or a pedestrian (c) crossing the road. But lo! somehow it passed on smoothly, leaving all obstacles (d) safely behind.
a) settlement
b) swallowing
c) walker
d) barriers

12. [delight, move slowly, meticulous, beeped, noise, ran]
Suddenly Valli clapped her hands with glee (a). A young cow, tail high in the air, was running very fast, right in the middle of the road, right in front of the bus. The bus slowed to a crawl (b), and the driver sounded his horn loudly again and again. But the more he honked (c), the more frightened the animal became and the faster it galloped (d) – always right in front of the bus.
a) delight
b) move slowly
c) beeped
d) ran

13. [outpost, dot, clatter, careful, huge, goods]
At last the cow moved off the road. And soon the bus came to a railroad crossing. A speck (a) of a train could be seen in the distance, growing bigger and bigger as it drew near. Then it rushed past the crossing gate with a tremendous (b) roar and rattle (c), shaking the bus. Then the bus went on and passed the train station. From there it traversed a busy, well-laid-out shopping street and, turning, entered a wider thoroughfare. Such big, bright-looking shops! What glittering displays of clothes and other merchandise (d)! Such big crowds! Struck dumb with wonder, Valli gaped at everything.
a) dot
b) huge
c) clatter
d)goods

Antonyms

Read the paragraph and match the words given in Column ‘A’ with the Antonyms in Column ‘B’

1. There was a girl named Valliammai who was called Valli for short (a). She was eight years old and very curious (b) about things. Her favourite pastime was standing in the front (c) doorway of her house, watching what was happening in the street outside. There were no playmates of her own (d) age on her street, and this was about all she had to do.

 A B a) short 1) difficult b) curious 2) improper c) front 3) disown d) own 4) back 5) apathetic 6) long

a-6, b-5, c-4, d-3

2. But for Valli, standing at the front door was every bit as enjoyable (a) as anv of the elaborate games other children played. Watching the street gave her many (b) new (c) unusual (d) experiences.

 A B a) enjoyable 1) few b) many 2) unenjoyable c) new 3) behind d) unusual 4) ended 5) usual 6) old

a-2, b-1, c-6, d-5

3. The most fascinating thing of all was the bus that travelled between her village and the nearest (a) town. It passed through her street each hour, once going to the town and once coming back. The sight of the bus, filled (b) each time with a new set of passengers, was a source of unending (c) joy (d) for Valli.

 A B a) nearest 1) ending b) filled 2) empty c) unending 3) farthest d) joy 4) sorrow 5) unordered 6) disclosed

a-3, b-2, c-1, d-4

4. Over many days and months Valli listened carefully (a) to conversations between her neighbours and people who regularly (b) used the bus, and she also asked a few discreet (c) questions here and there. This way she picked up various small details about the bus journey. The town was six miles from her village. The fare was thirty paise one way – “which is almost nothing at all,” she heard one well-dressed man say, but to Valli, who scarcely (d) saw that much money from one month to the next, it seemed a fortune.

 A B a) carefully 1) vacant b) regularly 2) distant c) discreet 3) abundantly d) scarcely 4) open 5) irregularly 6) carelessly

a-6, b-5, c-4, d-3

5. It was a new bus, its outside (a) painted a gleaming white with some green stripes along the sides. Inside, the overhead bars shone like silver. Directly in front of Valli, above the windshield, there was a beautiful (b) clock. The seats were soft (c) and luxurious (d).

 A B a) outside 1) frequently b) beautiful 2) spartan c) aoft 3) hard d) luxurious 4) ugly 5) inside 6) limited

a-5, b-4, c-3, d-2

6. Valli devoured (a) everything with her eyes. But when she started (b) to look outside, she found her view cut off by a canvas blind that covered (c) the lower part of, her window. So she stood up on the seat and peered (d) over the blind.

 A B a) devoured 1) ignored b) started 2) uncovered c) covered 3) ended d) peered 4) savored 5) simple 6) separate

a-4, b-3, c-2, d-1

7. The bus was now going along the bank of a canal. The road was very narrow (a). On one side there was the canal and, beyond (b) it, palm trees, grassland, distant (c) mountains, and the blue, blue sky. On the other side was a deep (d) ditch and then acres and acres of green fields – green, green, green, as far as the eye could see.

 A B a) narrow 1) near b) beyond 2) within c) distant 3) wide d) deep 4) abstained 5) revealed 6) shallow

a-3, b-2, c-1, d-6

8. Valli found the woman absolutely repulsive (a) – such big holes she had in her ear lobes, and such ugly (b) earrings in them! And she could smell the betel nut the woman was chewing and see the betel juice that was threatening (c) to spill over her lips at any moment.
Ugh! – who could be sociable (d) with such a person? “Yes, I’m travelling alone,” she answered curtly.

 A B a) repulsive 1) beautiful b) ugly 2) attractive c) threatening 3) within d) sociable 4) surface 5) antisocial 6) friendly

a-2, b-1, c-6, d-5

9. But the old woipan went on with her drivel. “Is it proper (a) for such a young (b) person to travel alone (c) ? Do you know exactly (d) where you’re going in town? What’s the street? What’s the house number?”

 A B a) proper 1) improper b) young 2) introverted c) alone 3) charming d) exactly 4) approximately 5) accompanied 6) old

a-1, b-6, c-5, d-4

10. Her first journey – what careful (a), painstaking, elaborate plans she had had to make for it! She had thriftily saved (b) whatever stray coins came her way, resisting (c) every temptation to buy peppermints, toys, balloons, and the like, and finally she had saved a total of sixty paise. How difficult (d) it had been, particularly that day at the village fair, but she had resolutely stifled a strong desire to ride the merry-go-round, even though she had the money.

 A B a) careful 1) careless b) saved 2) simple c) resisting 3) surrendering d) difficult 4) spent 5) together 6) antisocial

a-1, b-4, c-3, d-2

Right Forms of the Words

Fill in the blanks in the following passage choosing the right form of the words given in brackets. 4×1= 4M

1. There was a girl _____(a) (name / named / naming) Valliammai who was called Valli for short. She was eight years old and very curious about things. Her favourite pastime was _____(b) (stand / standing / stood) in the front doorway of her house, _____(c) (watch / watched / watching ) what was _____(d) (happen / happened / happening) in the street outside.
a) named
b) standing
c) watching
d) happening

2. The most _____(a) (fascinate / fascinated / fascinating) thing of all was the bus that travelled between her village and the _____(b) (near / nearer / nearest) town. It passed through her street each hour, once going to the town and once coming back. The sight of the bus, filled each time with a new set of passengers, was a source of unending joy for Valli. Day after day she watched the bus, and gradually a tiny wish crept into her head and grew there: she wanted to ride on that bus, even if just once. This wish became stronger and stronger, until it was an _____(c) (overwhelm / overwhelmed / overwhelming) desire. Valli would stare wistfully at the people who got on or off the bus when it stopped at the street corner. Their faces would kindle in her longings, dreams, and hopes. If one of her friends happened to ride the bus and tried to describe the sights of the town to her, Valli would be too jealous to listen and would shout, in English: “Proud! proud!” Neither she nor her friends really understood the meaning of the word, but they used it often as a slang _____(d) (express / expressing / expression) of disapproval.
a) fascinating
b) nearest
c) overwhelming
d) expression

3. Over many days and months Valli _____(a) (listen / listened / listening) carefully to conversations between her neighbours and people who regularly used the bus, and she also asked a few _____(b) (discreet / discreetly / discreetness) questions here and there. This way she picked up various small details about the bus journey. The town was six miles from her village. The fare was thirty paise one way – “which is almost nothing at all,” she heard one well-dressed man say, but to Valli, who _____(c) (scarce / scarcely / scarceness) saw that much money from one month to the next, it seemed a fortune. The trip to the town took forty-five minutes. On reaching town, if she _____(d) (stay / stayed / staying) in her seat and paid another thirty paise, she could return home on the same bus.
a) listened
b) discreet
c) scarcely
d) stayed

4. The conductor was a jolly sort, fond of _____(a) (joke / joked / joking). “Oh, please don’t be angry with me, my fine madam,” he said. “Here, have a seat right up there in front. Everybody move aside please-make way for madam. “It was the slack time of day, and there were only six or seven passengers on the bus. They were all looking at Valli and _____(b) (laugh / laughing / laughed) with the conductor. Valli was overcome with _____(c) (shy / shyly / shyness). Avoiding everyone’s eyes, she walked quickly to an _____(d) (empty / emptied / emptiness) seat and sat down.
a) joking
b) laughing
c) shyness
d) empty

5. It was a new bus, its outside _____(a) (paint / painted / painting) a _____(b) (gleam / gleamed / gleaming) white with some green stripes along the sides. Inside, the overhead bars shone like silver. Directly in front of Valli, above the windshield, there was a beautiful clock. The seats were soft and luxurious. Valli _____(c) (devour / devoured / devouring) everything with her eyes. But when she started to look outside, she found her view cut off by a canvas blind that _____(d) (cover / covered / covering) the lower part of her window. So she stood up on the seat and peered over the blind.
a) painted
b) gleaming
c) devoured
d) covered

6. The conductor _____(a) (punch / punched / punching) a ticket and handed it to her. “Just sit bafck and make yourself _____(b) (comfort / comforted / comfortable). Why should you stand when you’ve _____(c) (pay / paid / paying) for a seat?” “Because I want to,” she _____(d) (answer / answered / answering), standing up again.
a) punched
b) comfortable
c) paid

7. Valli found the woman absolute _____(a) (repulse / repulsion / repulsive) – such big holes she had in her ear lobes, and such ugly earrings in them! And she could _____(b) (smell / smelled / smelling) the betel nut the woman was _____(c) (chew / chewed / chewing) and see the betel juice that was _____(d) (threaten / threatened / threatening)to spill over her lips at any moment.
a) repulsive
b) smell
c) chewing
d) threatening

8. Her first journey – what careful, painstaking, _____(a) (elaborate / elaborated / elaboration) plans she had had to make for it! She had _____(b) (thrift / thriftily / thriftier) saved whatever stray coins came her way, _____(c) (resist /resisted / resisting) every _____(d) (tempt / tempted / temptation) to buy peppermints, toys, balloons, and the like, and finally she had saved a total of sixty paise.
a) elaborate
b) thriftily
c) resisting
d) temptation

9. After she had enough money saved, her next problem was how to slip out of the house without her mother’s knowledge. But she _____(a) (manage / managed / management) this without too much difficulty. Every day after lunch her mother would nap from about one to four or so. Valli always used these hours for her excursions’ as she stood _____(b) (look / looked / looking) from the doorway of her house or sometimes even _____(c) (venture / ventured / venturing) out into the village; today, these same hours could be _____(d) (use / used / using) for her first excursion outside the village.
a) managed
b) looking
c) ventured
d) used

10. The bus rolled on now _____(a) (cut / cuts / cutting) across a bare landscape, now _____(b) (rushes / rushed / rushing) through a tiny hamlet or past an odd wayside shop. Sometimes the bus seemed on the point of _____(c) (gobbles / gobbled / gobbling) up another vehicle that was coming towards them or a pedestrian crossing the road. But lo! somehow it _____(d) (pass / passed / passing) on smoothly, leaving all obstacles safely behind.
a) cutting
b) rushing
c) gobbling
d) passed

Vowel Clusters

Complete the spelling of the words choosing ‘ae’, ‘ee’, ‘ea’, ‘oo’, ‘ou’, ‘ai’ ‘ia’, ‘io’, ‘oi’, ‘au’ or ‘ua’.

Exercise – 1

1. She was eight y_ _rs (a) old and very cur_ _us (b) about things.
a) years
b) curious

2. Her fav_ _rite (a) pastime was standing in the front d_ _rway (b) of her house, watching what was happening in the street outside.
a) favourite
b) doorway

3. There were no playmates of her own age on her str_ _t (a), and this was ab_ _t (b) all she had to do.
a) street

4. Watching the street gave her many new unus_ _l (a) exper_ _nces (b).
a) unusual
b) experiences

5. The most fascinating thing of all was the bus that travelled between her village and the n_ _rest town.
nearest

6. The sight of the bus, filled each time with a new set of passengers, was a s_ _rce of unending joy for Valli.
source

7. Day after day she watched the bus, and grad_ _lly (a) a tiny wish crept into her h_ _d (b) and grew there: she wanted to ride on that bus, even if just once.

8. Their faces w_ _Id (a) kindle in her longings, dr_ _ms (b), and hopes.
a) would
b) dreams

9. Over many days and months Valli listened carefully to conversations between her neighb_ _rs (a) and people who regularly used the bus, and she also asked a few discr_ _t (b) questions here and there.
a) neighbours
b) discreet

10. This way she picked up vari_ _s (a) small details about the bus j_ _rney (b).
a) various
b) journey

Exercise – 2

1. On r_ _ching (a) town, if she stayed in her s_ _t (b) and p_ _d (c) another thirty p_ _se (d), she could return home on the same bus.
a) reaching
b) seat
c) paid
d) paise

2. Well, one fine spring day the aftern_ _n (a) bus was just on the p_ _nt (b) of leaving the village and turning into the main highway when a small v_ _ce (c) was heard shouting: “Stop the bus! Stop the bus!” And a tiny hand was r_ _sed (d) commandingly.
a) afternoon
b) point
c) voice
d) raised

3. “It’s me,” sh_ _ted Valli.
shouted

4. “Everybody move aside pi_ _se – make way for madam.” It was the slack time of day, and there were only six or seven passengers on the bus.

5. They were all l_ _king (a) at Valli and l_ _ghing (b) with the conductor.
a) looking
b) laughing

6. Av_ _ding (a) everyone’s eyes, she walked quickly to an empty s_ _t (b) and sat down.
a) Avoiding
b) seat

7. Then he blew his whistle twice, and the bus moved forward with a r_ _r.
roar

8. It was a new bus, its outside painted a gl_ _ming (a) white with some gr_ _n (b) stripes along the sides.
a) gleaming
b) green

9. Directly in front of Valli, above the windsh_ _Id (a), there was a be_ _tiful (b) clock.
a) windshield
b) beautiful

10. The seats were soft and luxur_ _us.
luxurious

Suffixes or Inflections

Fill in the blanks with correct suffixes given in the brackets.

Exercise – 1

1. There was a girl nam_____(ing / ed) Valliammai who was called Valli for short.
named

2. Her favourite pastime was standing in the front doorway of her house, watch_____(ed / ing) (a) what was happen_____(ed / ing) (b) in the street outside.
a) watching
b) happening

3. But for Valli, standing at the front door was every bit as enjoy _____(ing / able) as any of the elaborate games other children played.
enjoyable

4. The most fascinat_____(ed / ing) (a) thing of all was the bus that travelled between her village and the near_____(er / est) (b) town.
a) fascinating
b) nearest

5. The sights of the bus, fill_____(ing / ed) (a) each time with a new set of passengers, was a source of unend_____(ing / ed) (b) joy for Valli.
a) filled
b) unending

6. This wish became strong_____(er / est) (a) and stronger, until it was an overwhelm _____(ed / ing) (b) desire.
a) stronger
b) overwhelming

7. Neither she nor her friends really understood the meaning of the word, but they used it often as a slang expre_____(sion / ssion) (a) of disapprov_____(ed / al) (b).
a) expression
b) disapproval

8. Over many days and months Valli listened careful_____(ness/ly)(a) to conversa_____(sions / tions) (b) between her neighbours and people who regularly used the bus, and she also asked a few discreet questions here and there.
a) carefully
b) conversations

9. This way she pick_____(ing / ed) up various small details about the bus journey.
picked

10. The fare was thirty paise one way – “which is almost nothing at all,” she heard one well-dress_____(ing / ed) man say, but to Valli, who scarcely saw that much money from one month to the next, it seemed a fortune.
well-dressed

Exercise – 2

1. On reach_____(ed / ing) town, if she stayed in her seat and paid another thirty paise, she could return home on the same bus.
reaching

2. They were all look_____(ed / ing) (a) at Valli and laugh_____(ed / ing) (b) with the conductor.
a) looking
b) laughing

3. It was a new bus, its outside painted a gleam_____(ing / ed) white with some green stripes along the sides.
gleaming

4. It was an elderly man who had honestly been concerned for her, but she was annoy_____(ing / ed) (a) by his atten_____(tion / ssion) (b).
a) annoyed
b) attention

5. “Just sit back and make your_____(selves / self) (a) comfort_____(ness / able) (b).
a) yourself
b) comfortable

6. Valli found the woman absolute_____(ly / lly) (a) repulsive – such big holes she had in her ear lobes, and such ugly earrings in them! And she could smell the betel nut the woman was chewing and see the betel juice that was threaten_____(ed / ing) (b) to spill over her lips at any moment.
a) absolutely
b) threatening

7. Her first journey – what care_____(ful / full) (a), painstaking, elaborate plans she had had to make for it! She had thrifti_____(ness / ly) (b) saved whatever stray coins came her way, resisting every temptation to buy peppermints, toys, balloons, and the like, and finally she had saved a total of sixty paise.
a) careful
b) thriftily

8. “All by myself? Oh, I’d be much too afraid.” Greatly amus_____(ing / ed) (a) by the girl’s way of speak_____(er / ing) (b)
a) amused
b) speaking

9. The memory of the dead cow haunted her, dampen_____(ed / ing) her enthusiasm.
dampening

10. “What’s that you say?” “Oh,” said Valli, “I was just agreeing with what you Said about things happen_____(ed / ing) without our knowledge.”
happening

Identifying the Wrongly Spelt Word

Find the wrongly spelt word and write the correct spelling.

Exercise – 1

1. sensitive
2. fascinating
3. stretched
4. devoured
5. irritably
6. commandingly
7. concerned
8. pedestrian
9. traversed
10. enthusiasm

Exercise – 2

1. shrugged
2. chatterbox
3. overwhelming
4. disapproved
5. replanned
6. ventured
7. thirsty
8. absolutely
9. threatening
10. difficulty

Dictionary Skills

1. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘curt’?

b) What are the synonyms of the word ‘curt’?
abrupt; brusque

2. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘discreet’?

b) What is the synonym of ‘discreet’?
tactful

3. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘elaborate’?

b) What is the noun form of ‘elaborate’?
elaborateness

4. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the part of speech of the word ‘enjoyable’?

b) What is the adverb form of ‘enjoyable’?
enjoyably

5. Read the following dictionary entry of the word.

Now, answer the following questipns using the information above.
a) What are the other degrees of ‘haughty’?
haughty – haughtier – haughtiest

b) What is the synonym of the word ‘haughty’?
arrogant

6. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the synonym of ‘irritable’?

b) What is the noun form of ‘irritable’?
irritability

7. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the synonym of ‘luxurious’?
sumptuous

b) What is the opposite of ‘luxurious’?
spartan

8. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the synonym of ‘repulsive’?
disgusting

b) What is the adverb form of ‘repulsive’?
repulsively

9. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the synonym of ‘resolute’?
determined

b) What is the opposite of ‘resolute’?
irresolute

10. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What are the other degrees of comparison for ‘shy’?
shy – shyer – shyest

b) What is the synonym of ‘shy’?
timid

11. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the synonym of the word ‘sociable’?
gregarious

b) What is the opposite of ‘sociable’?
unsociable

12. Read the following dictionary entry of the word.

Now, answer the following questions using the information above.
a) What is the adverb form of ‘wistful’?
wistfully

b) What is the noun form of ‘wistful’?
wistfulness

Classification of Words

1. playmates, palm trees, mountains, passengers, grasslands, neighbours, conductor, fields

 People Scenery 1. 1. 2. 2. 3. 3. 4. 4.

 People Scenery 1. playmates 1. palm trees 2. passengers 2. mountains 3. neighbours 3. grasslands 4. conductor 4. fields

2. bank, hamlet, doorway, house, window, veranda, train station, thoroughfare

 Places Parts of the house 1. 1. 2. 2. 3. 3. 4. 4.

 Places Parts of the house 1. bank 1. doorway 2. hamlet 2. house 3. train station 3. window 4. thoroughfare 4. veranda

3. shop, village, bus, train, aeroplane, car, hamlet, street

 Places Vehicles 1. 1. 2. 2. 3. 3. 4. 4.

 Places Vehicles 1. shop 1. bus 2. village 2. train 3. hamlet 3. areoplane 4. street 4. car

4. head, earrings, coins, toys, earlobes, hands, eyes, ticket

 Parts of the body Things 1. 1. 2. 2. 3. 3. 4. 4.

 Parts of the body Things 1. head 1. earrings 2. earlobes 2. coins 3. hamlet 3. toys 4. eyes 4. ticket

Matchings

Match the words / phrases given under Column ‘A’ with their meanings given under Column ‘B’.

1.

 Column – A Column – B 1) inquisitive A) almost none 2) intricate B) extremely large or great 3) mesmerizing C) very interested about many different things 4) immense D) having a strong effect on you that you cannot give your attention to anything else E) having a lot of different parts and small details that fit together

1-C, 2-E, 3-D, 4-B

2.

 Column – A Column – B 1) turn off A) to stop working because of a fault 2) fill in B) to become very bad 3) break down C) to stop the flow of electricity, gas, water etc. 4) get off D) to fill something completely E) to leave / get down from F) to take someone’s place

1-C, 2-D, 3-A, 4-E

Section – C

Creative Expression

Conversation

Question 1.
From the lesson ‘Madam Rides the Bus’ you have explored Valli’s independence and determination to make her own decisions, even at a young age, such as saving money and planning her bus ride without her parents’ knowledge.
Now write a possible conversation between you and your elder brother on the importance of independence and seif realization.
I : Hey, brother, can we talk for a bit?
Brother : Of course, what’s on your mind?
I : Well, I’ve been thinking about how important it is to be independent and figure things out for yourself.
Brother : That’s a good thing to think about. Independence means being able to make your own choices and decisions.
I : Yeah, like Valli in that story we read, she wanted to ride the bus so badly, she saved money and planned everything herself, without even telling her parents.
Brother : Exactly. It shows determination and self-reliance. Independence isn’t just about doing things alone, it’s about knowing yourself and what you want.
I : I agree. It’s like learning to be responsible for your own actions and choices, right?
Brother : Absolutely. When you’re independent, you have the freedom to explore, make mistakes, and learn from them. It’s all about self¬realization, understanding who you are and what you’re capable of.
I : Yeah, and it helps us grow and become stronger individuals. I think it’s something we should all strive for.
Brother : Definitely. But remember, independence doesn’t mean shutting others out. It’s about finding a balance between relying on yourself and being open to help and guidance from others when you need it.
I : That makes sense. Thanks for talking about this with me. It’s good to know I can always count on you for advice.
Brother : Anytime, my dear one. Just keep believing in yourself and following your dreams. That’s what independence and self-realization are all about.

Question 2.
The poem ‘The Tale of Custard the Dragon’ itself is a story within a story. You could discuss the importance of storytelling as a means of conveying messages, entertaining audiences, and exploring themes and emotions.
Write a possible conversation between you and your friend Ramesh.
I : Hey Ramesh, have you ever read ‘The Tale of Custard the Dragon’?
Ramesh : No, I haven’t. What’s it about?
I : It’s a fun poem about a girl named Belinda and her unusual pets, including a cowardly dragon named Custard. It’s like a story within a story.
Ramesh : That sounds interesting. Why do you think stories like this are important?
I : Well, storytelling is a powerful way to share messages, entertain people, and explore different themes and emotions. It helps us understand complex ideas in simpler ways.
Ramesh : Yeah, I get that. Stories can really capture our imagination and teach us valuable lessons too.
I : Exactly! Plus, they bring people together. When we share stories, we connect with each other and learn about different perspectives and cultures.
Ramesh : I see what you mean. It’s like a way to understand the world around us better.
I : Definitely! That’s why storytelling has been around for centuries. Whether it’s through poems, books, movies, or even just sharing personal experiences, storytelling is a fundamental part of human communication.
Ramesh : I’m definitely going to check out ‘The Tale of Custard the Dragon’ now. Thanks for sharing!
I : No problem! 1 think you’ll enjoy it. Let me know what you think after you read it.

Question 3.
Analyze the relationship between Madame Loisel and Madame Forestier. How does Madame Forestier’s reaction to the lost necklace reflect on the nature of their friendship? How does betrayal impact relationships?
Write a possible conversation between you and your classmate Usha.
I : Hey Usha, do you remember the story we read, ‘The Necklace’?
I : Exactly! I found their relationship really interesting. Madame Loisel borrowed a necklace from Madame Forestier for the ball, but then she lost it.
Usha : Yeah, Madame Loisel had to replace the necklace because she lost it. It must have been really hard for her.
I : Definitely! But what struck me was Madame Forestier’s reaction when Madame Loisel finally confessed what happened. She seemed surprised and then revealed that the necklace was actually fake, worth much less than Madame Loisel thought.
Usha : That’s true. Madame Forestier’s reaction was unexpected. It’s like she didn’t really care about Madame Loisel’s struggles.
Usha : Yeah, it’s sad to see how betrayal can change a friendship. Madame Loisel went through so much because of that necklace, and Madame Forestier didn’t even seem to understand.
I : It shows how important trust and honesty are in friendships. If Madame Forestier had been more understanding, maybe Madame Loisel wouldn’t have suffered so much.
Usha : Definitely. It’s a good lesson for us to always be honest and supportive in our friendships.
I : Absolutely! It’s important to be there for each other, especially when things get tough. ,
Usha : Thanks for sharing your thoughts. It really helped me understand the story better.
I : No problem, Usha. I’m glad we could discuss it together!

Diary Entry

1. From the lesson ‘Madam Rides the Bus’ you have understood how Valli’s experiences on the bus ride shape her understanding of the world and her place in it.
Imagine yourself as Valli and write a Diary Entry in about 100 words
March 8, 20xx
Friday, 9.30 p.m.
Dear Diary,
Today was the most exhilarating day of my life! I finally rode the bus to town all by myself. The sights, the sounds, the people-everything felt so vibrant and alive. From the gleaming white bus to the bustling streets of the town, I soaked in every moment with wide-eyed wonder. But amidst the excitement, I couldn’t shake off the image of the dead cow on the roadside. It made me realize the fragility of life and the importance of cherishing every moment. This bus ride has shown me that the world is vast and full of mysteries waiting to be explored.
Until tomorrow,
Valli

Question 2.
Belinda and her companions, despite their differences and fears, support each other in times of danger. Now write a Diary Entry about your friendship in about 100 words.
March 8, 20xx
Friday, 9.30 p.m.
Today has been a day filled with reflections on friendship, courage, and support. As I sit down to pen my thoughts, I find myself overwhelmed by the warmth and strength that my companions have provided me, much like Belinda and her unusual pets in “The Tale of Custard the Dragon”.

In this journey called life, I’ve come to realize the profound significance of friendship. It’s not merely about having someone to share laughter and joy with, but it’s about having a steadfast support system in times of uncertainty and danger.

Just like Belinda, 1 am blessed with companions who, despite their differences and fears, stand by my side with unwavering loyalty and courage. Each of them brings a unique essence to our friendship, much like Ink, Blink, Mustard, and Custard in the poem.

In times of peril, their presence is my anchor. Their encouragement fuels my spirit, their laughter soothes my soul, and their unwavering belief in me strengthens my resolve. Together, we navigate the turbulent seas of life, facing challenges head-on, and celebrating victories with unbridled joy.

As I close this diary entry, I am reminded of the profound words of Helen Keller: “Walking with a friend in the dark is better than walking alone in the light.” Indeed, in the journey of life, I am grateful for the light that my friends bring, illuminating my path with their love, laughter, and unwavering support.
Until next time,
xxxx

Question 3.
You have read the lesson ‘The Necklace’ today. Now write a Diary Entry on the moral lessons that can be drawn from the story. What insights can be gained about the nature of happiness, the value of honesty, and the importance of gratitude and contentment?
February 14, 20xx
Wednesday, 9.30 p.m.
Dear Diary,
Today, I read the story ‘The Necklace’ and it left me with deep reflections on life and its lessons. The story revolves around Madame Loisel and her pursuit of wealth, status, and happiness, which ultimately leads to her downfall. Through Madame Loisel’s journey, several moral lessons emerge, offering insights into the nature of happiness, the value of honesty, and the importance of gratitude and contentment.

Firstly, the story underscores the fleeting nature of happiness derived from material possessions. Madame Loisel’s relentless desire for luxury and social status blinds her to the simple joys of life. She learns the hard way that true happiness does not stem from wealth or material possessions but from inner contentment and appreciation for what one has.

Moreover, the story highlights the importance of honesty and integrity in relationships. Madame Loisel’s decision to deceive Madame Forestier about the lost necklace leads to years of hardship and suffering. It serves as a reminder that honesty is the foundation of trust and genuine connections with others. Without honesty, relationships crumble under the weight of deceit and mistrust.

Furthermore, ‘The Necklace’ emphasizes the significance of gratitude and contentment in finding fulfillment in life. Madame Loisel’s inability to appreciate her modest lifestyle and the love of her husband leads to a life consumed by regret and hardship. It teaches us the importance of being grateful for the blessings we have and finding contentment in the present moment.

Today’s reading has left a lasting impression on me, prompting me to reevaluate my priorities and strive for a life guided by authenticity, gratitude, and inner peace.
Until next time,
xxxx

Letter Writing

Question 1.
Voting in elections is crucial for democracy. It gives citizens a voice in choosing leaders and shaping policies. Every vote counts, influencing who governs and the direction of society. Through voting, individuals express their preferences and hold leaders accountable. Voting is a fundamental right and responsibility that empowers citizens to contribute to their communities and shape the future.
Draft a letter to a friend explaining the importance of voting in elections.

 Narasaraopet, 208.03.20xx. Dear [Friend’s Name], I hope this letter finds you well. As we navigate through the complexities of our ’ society, I find myself reflecting on the fundamental principles that shape our democracy. One such principle, which I believe is crucial to our collective well-being, is the act of voting in elections. Voting, as we both know, is far more than just marking a ballot or casting a preference. It embodies the very essence of democracy itself. It empowers individuals like you and me to have a say in who represents us, the policies they enact, and the direction our society takes. Moreover, voting is not merely a privilege but a duty-one that demands our active engagement and commitment. Our democracy thrives when citizens like us actively participate, informed by our values, beliefs, and aspirations for a better future. It’s a reflection of our commitment to our communities, our country, and the ideals we hold dear. I understand that amidst the busyness of life, it’s easy to overlook the significance of voting. Yet, I urge you to consider the profound impact each vote carries-the potential to shape policies, safeguard freedoms, and champion the causes we hold dear. Your voice matters, my friend, and your vote is a testament to your commitment to a better tomorrow. As we approach the upcoming elections, I encourage you to reflect on the values that drive you, the issues that resonate with you, and the future you envision for our society. Let us not squander the opportunity to make a difference, however small it may seem. Together, through our collective voices and actions, we can shape a brighter, more inclusive future for generations to come. With warm regards, [Your Name]

Question 2.
Extracurricular activities at school are vital for holistic development. They offer students opportunities beyond academics to explore interests, build skills, and develop social bonds. Participation in sports, clubs, and arts cultivates teamwork, leadership, and time management. These activities nurture creativity, critical thinking, and problem¬solving abilities, preparing students for real-world challenges.
Pen a letter to your school principal proposing a new extracurricular activity.
[City, State, ZIP Code]
[Phone Number]
[Date]
[Principal’s Name]
[School Name]
[City, State, ZIP Code]
Dear [Principal’s Name],
I hope this letter finds you well. As a student at [School Name], I have come to deeply appreciate the value and importance of extracurricular activities in shaping our holistic development. These activities not only provide avenues for exploration and growth but also foster a sense of community and belonging among students.

I am writing to propose the introduction of a new extracurricular activity that I believe will greatly benefit our school community: a Debate Club.

The establishment of a Debate Club would offer students a platform to engage in structured debates, where they can articulate their opinions, defend their viewpoints, and engage in critical thinking. Through participation in debates, students can refine their communication skills, develop the ability to think on their feet, and learn the art of persuasive argumentation.

I am willing to take the initiative in organizing and coordinating the activities of the Debate Club, with the support and guidance of the school administration and faculty. I believe that the Debate Club will not only enrich the extracurricular landscape of our school but also contribute to the academic and personal growth of its members.

I kindly request your consideration of this proposal and would be more than happy to discuss it further at your convenience. Thank you for your attention to this matter, and 1 look forward to the opportunity to contribute to the enrichment of our school community.
Warm regards,

Question 3.
Eco-friendly initiatives are essential for preserving our planet. These initiatives aim to reduce environmental impact by promoting sustainable practices and renewable resources. They encourage recycling, energy conservation, and reducing carbon emissions. Eco-friendly initiatives advocate for protecting biodiversity and natural habitats while minimizing pollution and waste. From renewable energy projects to green transportation solutions, these efforts seek to create a healthier, cleaner future for generations to come.
Draft a letter to a local environmental organization suggesting eco-friendly initiatives.
[City, State, ZIP Code]
[Phone Number]
[Date]
[Organization Name]
[City, State, ZIP Code]
Dear [Organization Name],
I hope this letter finds you well. As a passionate advocate for environmental conservation and sustainability, I am writing to share some ideas and suggestions for eco-friendly initiatives that could be implemented in our local community.

In recent years, there has been a growing awareness of the urgent need to address environmental issues and reduce our collective carbon footprint. As a result, I believe there is an opportunity for our community to take proactive steps towards promoting sustainability and preserving our precious natural resources.

Here are a few suggestions for eco-friendly initiatives that I believe could make a positive impact in our community:
1. Community-wide Recycling Programme :
Implementing a comprehensive recycling programme that encourages residents and businesses to recycle materials such as paper, glass, plastic, and metal can significantly reduce waste sent to landfills and promote a culture of recycling.

2. Local Environmental Education Programmes:
Organize workshops, seminars, and educational events to raise awareness about environmental issues and promote sustainable living practices within our community. Topics could include composting, water conservation, sustainable gardening, and reducing single-use plastics.

3. Community Clean-up Events :
Coordinate regular clean-up events in local parks, beaches, and natural areas to remove litter and debris, and restore the beauty of our natural surroundings. Engaging volunteers of all ages in hands-on conservation activities fosters a sense of community pride and environmental stewardship.

I am eager to collaborate with your organization and other community stakeholders to explore the feasibility of implementing these eco-friendly initiatives in our local area. Together, we can work towards building a more sustainable and resilient community that prioritizes the well-being of our planet and future generations.

Thank you for considering these suggestions, and I look forward to the opportunity to contribute to our shared vision of a healthier, cleaner environment.
Warm regards,

Biographical Sketch

Question 1.
Write a biographical sketch of Dr. A.P.J. Abdul Kalam on the basis of the notes given below.
Name : A.P.J. Abdul Kalam
Popularly known as : Missile Man
Famous as : 11th President of India
Born on : 15.10.1931
Place of Birth : Rameswaram, Tamil Nadu
Died on : 27.07.2015, in Shillong, Meghalaya
Parents : Jainulabdeen (Father), Ashiamma (Mother)
Awards : Bharat Ratna (1997),
Achievements : Evolution of ISRO’s launch vehicle programme, operationalisation of AGN1, PRITHVI missiles.
Literary pursuits : Wings of Fire
– India 2020 – A Vision for the New Millennium
– My Journey
– Ignited Minds.
A.P.J. Abdul Kalam, the 11th President of India, is popularly known as the ‘Missile Man’ of India. His full name is Avul Pakir Jainulabdeen Abdul Kalam. He was a famous Indian scientist. He was born on October 15th, 1931 at Rameswaram, Tamil Nadu. His parents were Jainulabdeen and Ashiamma. He played a key role in evolution of ISRO’s launch vehicle programme and operationalisation of AGN1 and PRITHVI missiles. Wings of Fire, India 2020 – A Vision for the New Millennium, My Journey and Ignited Minds were some of his literary pursuits. He was awarded Bharat Ratna in 1997, Padma Vibhushan in 1990 and Padma Bhushan in 1981. He died on 27th July, 2015 in Shillong of Meghalaya.

Question 2.
Write a biographical sketch of Mahanati Savitri using the information given below.
Born : December 6, 1937, Guntur District
Spouse : Gemini Ganesan
Children : Vijaya Chamundeswari, Satish Kumar Ganesan
Debut Film : Agnipareeksha
Career : Nearly 300 films; 30 year long; Acted in Telugu, Tamil, Kannada and Hindi films.
Milestone in her career : Chivaraku Migiledi
Memorable Performances : Missamma, Devadasu, Maya Bazaar, etc.
Honours : Mahanati, Kalaimamani, Nadigayar Thilakam titles.
Awards : Presidential award for her performance in ‘Chivaraku Migiledi’
Died : December 26, 1981, Chennai.
Savitri was a great actress of Indian film industry. She was called Mahanati. She was born on December 6, in 1937 in Guntur District. She had her schooling in Vijayawada. She completed her third form. She married Gemini Ganesan and had two children, Vijaya Chamundeswari and Satish Kumar Ganesan. Agnipareeksha was her debut film. She performed nearly in 300 films and she had a 30 year long career in film industry. She acted in Telugu, Tamil, Kannada and Hindi films. Chivaraku Migiledi was a mile¬stone in her career. Missamma, Devadasu and Maya Bazaar were her great movies. She received the honours Mahanati, Kalaimamani and Nadigayar Thilakam titles. She received the Presidential award for her performance in ‘Chivaraku Migiledi. She passed away on December 26 in 1981 in Chennai.

Question 3.
Write a biographical sketch of Asha Parekh, a famous Indian actress, film director, producer and the most successful actress in Hindi cinema.
Name : Asha Parekh
Date of Birth : 2 October 1942
Place of birth : Bombay
Occupation : Actress, Director and Producer
Father : Bachubhai Parekh
Mother : Salma Parekh
First film : Maa
Credit : Highest paid actress of her time
Movie which made her a star : Dil Deke Dekho
Asha Parekh is a famous Indian actress, film director, producer and the most successful actress in Hindi cinema. She was born on 2 October 1942 in Bombay. Her father is Bachubhai Parekh and mother, Salma Parekh. Her first film was ‘Maa’ which was released in 1952. She started her career as a child artist acting in ‘Maa’ at the age of 10. The film Dil Deke Dekho (1959) made her a star heroine. She was the highest paid actress of her time. She was honoured with the Padma Shri Award in 1992 by the government of India for her contribution to the field of cinema. She was also honoured with the Dada Saheb Phalke Award in 2020.

Framing ‘Wh’ Questions

Read the following passage carefully focussing on the underlined parts.

1. She was one of those pretty, voung ladies (A) born as if through an error of destiny (B) into a family of clerks. She had no dowry, no hopes, no means of becoming known (C), loved, and married by a man either rich or distinguished (D); and she allowed herself to marry a petty clerk in the office of the Board of Education (E).
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) What kind of a woman was she?
B) How was she born?
C) What did she not have?
D) What kind of a man loved and married her?
E) What did she allow herself to do?

2. One evening her husband returned elated (A) bearing in his hand a large envelope (B). “Here,” he said, “here is something for you.”
She quickly drew out a printed card (C) on which were inscribed these words :
The Minister of Public Instruction and Madame George Ramponneau (D) ask the honour of M. and Mme. Loisel’s (E) company, Monday evening, January 18, at the Minister’s residence.
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) How did her husband return one evening?
B) What was he bearing in his hand?
C) What did he quickly draw out?
D) Who were the people inviting?
E) Who were being invited?

3. He was silent, stupefied, in dismay (A) at the sight of his wife weeping (B). He stammered, “What is the matter? What is the matter?”
By a violent effort (C), she had controlled her vexation (D) and responded in a calm voice (E), wiping her moist cheeks.
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) How was he?
B) What was he silent and stupefied at?
C) How had she controlled her vexation?
E) How did he respond?

4. The day of the ball (A) approached and Mme Loisel seemed sad, disturbed, anxious (B). Neverthless her dress (C) was nearly ready. Her husband said to her one evening, “What is the matter with you ? You have acted strangely (D) for two or three days” (E).
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) What day approached?
B) How did Mme Loisel seem?
D) How has she acted?
E) How long has she acted strangely?

5. The next day she took herself to her friend’s house (A) and she related her story of distress (B) Mme Forestier (C) went to her closet, took out a large iewel-case (D), brought it, opened it, (E) and said, “Choose, my dear.”
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.
A) Where did she take herself to?
B) What did she relate?
C) Who went to her closet?
D) What did she take out?
E) What did she do?

6. The day of the ball (A) arrived. Mme Loisel (B) was a great success. She was the prettiest of all – elegant, gracious, smiling and full of joy (C). All the men (D) noticed her, asked her name and wanted to be presented. She danced with enthusiasm- intoxicated with pleasure (E).
Now frame ‘WH’ questions to get the underlined parts in the passage as answers.