Class 9

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes InText Questions

Try This

Take a cube of edge T cm and cut it as we did in the previous activity and find total surface area and lateral surface area of cube. [Page No. 216]
Solution:

If we cut and open a cube of edgewe obtain a figure as shown above.
In the figure, A, B, C, D, E, F are squares of side
The faces A, C, D, F forms the lateral surfaces of the cube.
∴ Lateral surface area of the cube = 4l²
And all six faces form the cube.
∴ Total surface area of the cube = 6l²

Do This

Question 1.
Find the total surface area and lateral surface area of the cube with side 4 cm. By using the formulae deduced in above. Try this. [Page No. 216]1. Find the total surface area and lateral surface area of the cube with side 4 cm. By using the formulae deduced in above. Try this. [Page No. 216]

Solution:
Total surface area of a cube = 6l²
Given l = 4crn
∴ T.S.A. = 6 × 4² = 6 × 16 = 96 cm²
∴ L.S.A. = 4l² = 4 × 4² = 64cm²

Question 2.
Each edge of a cube is increased by 50%. Find the percentage increase in the surface area. [Page No. 216]
Solution:
Let the edge of a cube x units
Its surface area = 6l² = 6x² sq. units
If its edge is increased by 50%, then
new edge = x + 50% of x


Where x = increase/decrease

Try These

Question
a) Find the volume of a cube whose edge is ‘a’ units. [Page No. 217]

Solution:
V = edge³ = a³ cubic units.

b) Find the edge of a cube whose volume is 1000 cm³. [Page No. 217]
Solution:
V = edge³ = 1000 = 10 × 10 × 10 = 10³
∴ Edge =10 cm

Do These

Question 1.
Find the volume of a cuboid if l = 12 cm, b = 10 cm, h = 8 cm. [Page No. 218]
Solution:
Volume V = lbh = 12 × 10 × 8 = 960cm³

Question 2.
Find the volume of cube, if its edge is 10 cm [Page No. 218]
Solution:
Volume V = l³ = 10 × 10 × 10 = 1000cm³

Question 3.
Find the volume of isosceles right angled triangular prism. [Page No. 218]

Solution:
Volume = Area of base × height
= Area of isosceles triangle × height

Activity

Question
Take the square pyramid and cube containers of same base and with equal heights. [Page No. 218]

Fill the pyramid with a liquid and pour into the cube (prism) completely. How many times it takes to fill the cube? From this, what inference can you make?

Thus volume of pyramid = \(\frac { 1 }{ 3 }\) × of the volume of right prism
= \(\frac { 1 }{ 3 }\) × Area of the base × height.

Note : A right prism has bases perpendicu¬lar to the lateral edges and all lateral faces are rectangles.

Do These

Question 1.
Find the volume of a pyramid whose square base is 10 cm and height is 8 cm. (Page No. 219)
Solution:
Volume of a pyramid
= \(\frac { 1 }{ 3 }\) × Area of the base × height
= \(\frac { 1 }{ 3 }\) × 10 x 10 × 8 = \(\frac { 800 }{ 3 }\) cm³

Question 2.
The volume of a cube is 1200 cubic cm. Find the volume of square pyra¬mid of the same height. (Page No. 219)
Solution:
Volume of the square pyramid
= \(\frac { 1 }{ 3 }\) × volume of the square prism
= \(\frac { 1 }{ 3 }\) × 1200 = 400 cm³

Activity

Cut out a rectangular sheet of paper. Paste a thick string along the line as shown in the figure. Hold the string with your hands on either sides of the rectangle and rotate the rectangle sheet about the string as fast as you can.

Do you recognize the shape that the rotating rectangle is forming ?
Does it remind you the shape of a cylinder ? [Page No. 220]

Do This

Question 1.
Find C.S.A. of each of the following cylinders. [Page No. 221]
i) r = x cm; h = y cm
Solution:
CSA = 2πrh = 2πxy cm²

ii) d = 7 cm; h = 10 cm
Solution:

CSA = 2πrh
= 2 × \(\frac{22}{7} \times \frac{7}{2}\) × 10
= 220 cm²

iii) r = 3 cm; h = 14 cm
Solution:

CSA = 2πrh
= 2 × \(\frac{22}{7} \times \frac{7}{2}\) × 3 × 14
= 264 cm²

Question 2.
Find the total surface area of each of the following cylinder. [Page No. 222]
i)

Solution:
r = 7 cm; h = 10 cm
T.S.A. = 2πr (r + h)
=2 × \(\frac{22}{7}\) × 10 (7 + 10)
= 2 × \(\frac{22}{7}\) × 10 × 17
= 1068.5 cm²

ii)

h = 7cm; πr² = 250
πr² = 250 = 250
\(\frac{22}{7}\) × r² = 250 = 250
r² = 125 x \(\frac{7}{11}\)

Try These

Question 1.
If the radius of a cylinder is doubled keeping Its lateral surface area the same, then what is its height? [Page No. 225]
Solution:
Let the initial radius and height of the
cylinder be r and h.
Then L.S.A. = 2πrh
When r is doubled and the L.S.A. remains the same, then the height be hr By problem new L.S.A. = 2πrh
= 2π (2r) (h₁)
⇒ 2πrh = 4πrh₁
∴ \(\frac{2 \pi \mathrm{rh}}{4 \pi \mathrm{r}}=\frac{1}{2} \mathrm{~h}\)
Height becomes its half.

Question 2.
A hot water system (Geyser) consists of a cylindrical pipe of length 14 m and diameter 5 cm. Find the total radiating surface of hot water system. [Page No. 225]
Solution:
Radius (r) = \(\frac{\text { diameter }}{2}=\frac{5}{2}\) = 2.5 cm
Length of the pipe = height = 14 m Radiating surface = 2πrh
= 2 × \(\frac{22}{7}\) × 2.5 × 1400
= 22000 cm³

Activity

Question
Making a cone from a sector. [Page No. 227] Follow the instructions and do as shown in the figure.

i) Draw a circle on a thick paper Fig (a).
ii) Cut a sector AOB from it Fig (b).
iii) Fold the ends A, B nearer to each other slowly and join AB. Remember A, B should not overlap on each other. After joining A, B attach them with cello tape Fig (c).
iv) What kind of shape you have obtained? Is it a right cone?
While making a cone observe what hap-pened to the edges ‘OA’ and ‘OB’ and length of arc AB of the sector?

Try This

A sector with radius r and length of its arc / is cut from a circular sheet of paper. Fold it as a cone. How can you derive the formula of its curved surface area A = πrl. [Page No. 228]

C.S.A = πrl
Solution:
When a sector of radius ‘r’ and whose length of arc l is folded to form a cone. Radius ‘r’ becomes slant height ‘l’ and arc ‘l’ becomes perimeter of the base 2πr.

∴ Area of the sector = \(\frac{l r}{2}\) = Area of the cone
\(\frac{2 \pi \mathrm{r} l}{2}\) = Surface area of the cone
C.S.A = πrl

Do This

Question 1.
Cut a right angled triangle. Stick a string along its perpendicular side, as shown in fig. (T) hold both the sides of a string with your hands and rotate it with constant speed. What do you observe ? [Page No. 229]

Solution:
A right circular cone is observed.

Question 2.
Find the curved surface area and total surface area of the each following right circular cones.[Page No. 229]
Solution:

OP = 2 cm; OB = 3.5 cm
OP = h = 2 cm

OP = 3.5 cm; AB = 10 cm
r = \(\frac{\mathrm{AB}}{2}\) = 5cm; h = 3.5cm
r = OB = 3.5 cm
C.S.A. = πrl

T.S.A. = πr (r + l)
= \(\frac{22}{7}\) × 3.5(3.5 + 4.03)
= \(\frac{22}{7}\) × 3.5 × 7.53 = 82.83cm²

C.S.A. = πrl
\(\frac{22}{7}\) × 5 × 6.10
= 95.90cm²

T.S.A. = πr (r + l)
= \(\frac{22}{7}\) × 5 × (5 + 6.10)
= 174.42 cm²

Activity

Draw a circle on a thick paper and cut it neatly. Stick a string along its diam¬eter. Hold the both the ends of the string with hands and rotate with con¬stant speed and observe the figure so formed. [Paper 235]

Try This

Question 1.
Can you find the surface area of sphere in any other way ? (Page No. 235)
Solution:

Height of the pyramid is equal to r.
To derive the formula of the surface area of a sphere, we imagine a sphere with many pyramids inside of it until the base of all the pyramids cover the entire surface area of the sphere. In the figure below, only one of such pyra¬mid is shown.
Then, do a ratio of the area of the pyramid to the volume of the pyramid.
The area of the.pyramid is A.
The volume of the pyramid is V = (1/3) × A × r = (A × r)/3
So, the ratio of area to volume is A/V = A + (A × r) / 3 = (3 × A) / (A × r) = 3 / r

For a large number of pyramids, let say that n is such large number, the ratio of the surface area of the sphere to the volume of the sphere is the same as 3/r.
For n pyramids, the total area is n × A- Also for n pyramids, the total volume is n × V.
Therefore, ratio of total area to total volume is n × A/n × V = A/ V and we already saw before that A / V = 3 / r

Further more, n × A_pyramid = A_sphere (The total area of the bases of all pyramids or n pyramids is approximately equal to the surface area of the sphere).
n × V_pyramid = V_sphere (The total Volume of all pyramids or n pyramids is approximately equal to the volume of the sphere.
Putting observation # 1 and # 2 together, we get

Therefore, the total surface area of a sphere, call it S.A. is 4πr ²

Do These

Question 1.
A right circular cylinder just encloses a sphere of radius r (see figure). Find i) surface area of the sphere
ii) curved surface area of the cylinder
iii) ratio of the areas obtained in (i) and (ii) [Page No. 236]

[Page No. 236]
Solution:
i) Radius of the sphere = radius of the cylinder = r
∴Surface area of the sphere = 4πr²
ii) C.S.A. of cylinder = 2πr (2r) [∵ h = 2r]
= 4πr²
iii) Ratio of (i) and (ii) = 4πr² : 4πr² = 1:1

Question 2.
Find the surface area of each of the following figure.
i)

Surface area = 4πr²
C.S.A = 4 × \(\frac{22}{7}\) × 7 × 7
= 616cm²

ii)

C.S.A = 2πr² = 2 × \(\frac{22}{7}\) × 7 × 7 = 308cm²
Total Surface area = 3πr²
= 3 × \(\frac{22}{7}\) × 7 × 7 = = 462cm²

Do This

Question 1.
Find the volume of the sphere given below[Page No. 238]

Solution:
r = 3 cm
V = \(\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7}\) × 3 × 3 × 3
= 113.14cm³

d = 5.4 cm
r = \(\frac{\mathrm{d}}{2}=\frac{5.4}{2}\) = 2.7 cm
V = \(\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7}\) × 2.7 × 2.7 × 2.7 = 82.48cm³

Question 2.
Find the volume of sphere of radius 6.3 cm. [Page No. 238]
Solution:
r = 6.3 cm
V = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7}\) × 6.3 × 6.3 × 6.3 = 1047.81cm³

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 3B Not Just a Teacher, but a Friend Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 3B Not Just a Teacher, but a Friend

9th Class English Chapter 3B Not Just a Teacher, but a Friend Textbook Questions and Answers

I. Answer the following questions.

Question 1.
Why does the student consider the teacher his friend?
Answer:
The student is sad. He has no one to share his sorrow. He doesn’t know where to go. Then, he meets his teacher. The teacher’s company makes him comfortable. Hence the boy consideres the teacher his friend.

Question 2.
How does the teacher take the student out of his gloom?
Answer:
The teacher listens to the boy. He offers the boy comfort. He provides the boy with the needed courage. He brings out the boy’s real soul. He finds time to see if the boy is fine. In the company of this teacher-friend, the boy is all smiles.

Question 3.
Why does the student fail to estimate the teacher correctly?
Answer:
The boy moves to the higher class. He is worried that he would miss his teacher-friend. As a young boy, he fails to understand that great teacher’s concern for him.

Question 4.
How does the poet glorify the relationship between the teacher and the student?
Answer:
The poet lays bare the boy’s heart. That brings out the dizzy heights of the teacher’s personality. The boy is sad. The teacher makes him glad. The boy feels lonely. The teacher makes him lively. The relationship between them is that of love and affection. They become a model teacher student pair. The bond between them makes everyone fond of them.

Not Just a Teacher, but a Friend Summary in English

The poem is a moving account of a noble teacher being a kind friend of a student with no one to depend on. The boy was sad. He had not a soul to share his sorrow with. He had nowhere to go. Then he saw this teacher. He trusted the teacher. He poured out his feelings. The teacher extended his healing hand. The boy felt relieved. Whenever the boy felt unhappy, he would go to his teacher and found comfort in his company.

When the boy moved to a higher grade, he was worried that he would miss his favourite teacher. But he was wrong. The relationship only grew strong. The teacher would find time to greet the boy. The boy would smile and feel happy. Hence the boy feels that the teacher is also his friend.

Not Just a Teacher, but a Friend Glossary

trust (v) : believe in ; have faith in

true soul (phrase) : the real self : the hidden power

next grade (phrase) : higher class

fade (v) : become dull; pale

fate (v) : destiny; future

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 3A Swami is Expelled from School Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 3A Swami is Expelled from School

9th Class English Chapter 3A Swami is Expelled from School Textbook Questions and Answers

Look at the picture and answer the questions that follow.

Question 1.
Why do you think the teacher is punishing that student?
Do you approve of this action of the teacher? Give at least one reason for your opinion.
Answer:
I think the teacher is punishing the student because he might cause nuisance. He might not finish his homework. He might be indisciplined in the class. I usually don’t approve this kind of action of the teacher because corporal punishment should not be initiated against children. It is common with the school children behaving improperly at their tender age. They don’t know the importance of education and how to behave. Teachers need great patience to change them.

Question 2.
Can you suggest a few steps to correct the students causing nuisance?
Answer:
Various different kinds of steps are at hand to correct erring students. Some of them are:

1. explaining to them in the way they appreciate what is wrong with them and how they can be good.
2. making them imagine that they are at the receiving end.
3. making them realise the value of the fruits of being good.
4. providing them with examples of bad behaviour ruining careers and lives.
5. engaging them in productive activities of their choice.
6. repeatedly attempting to understand their feelings, sentiments, and view points.
7. giving them opportunities to express themselves, to lead, to organise, to suggest programmes and activities.
8. ignoring their misdeeds and allowing and encouraging them to correct themselves.
9. informing their parents about their positive qualities.
10. encouraging them to participate in social service activities.

Comprehension

I. Answer the following questions.

Question 1.
Why do you think the headmaster entered the class with a flushed face and a hard ominous look?
Answer:
Swami and his friends had not attended the classes the previous day. Moreover they had broken the panes of ventilators of the headmaster’s room. The school peon saw that. And he Informed the headmaster. That was why the headmaster was very angry.

Question 2.
Why did the headmaster send for the peon?
Answer:
The headmaster sent for the peon. He wanted to show to the students the proof of their mischief. The peon was the witness to their wrong deeds.

Question 3.
“I don’t care for your dirty school.” Why did Swami mutter so?
Answer:
The headmaster went on beating Swami. Swami complained of severe pain. That Invited four more hits. Swami could no longer bear the pain. He was in a desperate situation, That made him bold. So he took his books and walked out of the class. He said he least cared for that dirty school,

Question 4.
Do you justify the headmaster’s behaviour? If not, state your reasons.
Answer:
No, I don’t justify the headmaster’s behaviour. Corporal punishment is not the correct way to correct the wrong doing. And continuous corporal punishment is the worst part of it. That only provokes revolt not remorse.

Question 5.
If you were in Swami’s place, how would you feel?
Answer:
If I were in Swami’s place, I would feel the same way as Swami did. But I doubt I would have broken the panes. And I am afraid I could have muttered as Swami did.

II. Here are some utterances from the story. Complete the table.

Answer:

III. Work in Groups

Give reasons for Swami’s decision to leave the school:
1. ________________
2. ________________
3. ________________
Answer:

1. Swami was unable to bear the pain of the cane any more.
2. The headmaster had any how announced his dismissal.
3. More than the physical pain, the insult was unbearable.
4. In the desperation, Swami couldn’t think of any other alternative.
5. Swami must have thought of joining another school.

Vocabulary

I. Pick out words from the story which are synonyms of the following words.

Word	Synonyms
1. beat
2. angry
3. rascal
4. humiliation

Answer:

Word	Synonyms
1. beat	thrash, whack, rap
2. angry	flushed ; furious
3. rascal	loafer, idiot, cheat
4. humiliation	insult

II. Look at the following phrasal verbs taken from the text.

1. keep away
2. look around
3. look at
4. bring down
5. cut off

These phrases are verbs followed by prepositions or adverbial particles. You may understand that they are phrasal verbs.

You will notice that the following phrasal verbs ‘keep away’, ‘bring down’ and ‘cut off’ can be split as shown below.
Examples:

1. Keep the files away.
2. Bring the patient down.
3. Cut it off.

The other two phrasal verbs cannot be split.
Use the following phrasal verbs in your own sentences and decide whether you can split them as shown in the above examples.

Phrasal Verbs	Sentences of your own
look up
bring out
throw out
look out

Answer:

III. Refer to a dictionary and pick out the phrasal verbs that begin with the following verbs and write down sentences using them.

Verbs	Phrasal verbs beginning with the verb
bring
look
rush
keep
go
put

Grammar

I. Read the following imaginary conversation between Swami and the headmaster.

Headmaster : Why didn’t you come to school yesterday?
Swami : Sir, my mother was suffering from fever.
Headmaster : I don’t believe your words. You always say something or the other to escape from school.
Swami : I’m speaking the truth, sir.
Headmaster : Well, I’ll come to your house tomorrow and talk to your parents.

In the indirect speech, the above conversation can be written like this.

The headmaster asked Swami why he had not come to school the day before. Swami replied respectfully that his mother had been suffering from fever. The headmaster retorted that he did not believe his words and added that he always said something or the other to escape from school. Then Swami replied respectfully that he was speaking the truth. Then the headmaster told him that he would go to his house the next day and talk to his parents.

As you can see, while converting the direct speech into indirect speech, the words in the bold are added to express the feelings, emotions, attitudes of the speaker, and the sequences of the actions.

Read the imaginary conversation between Swami and his father:

Swami’s Father : My dear Swami, why are you looking so dull? Why haven’t you gone to school today?
Swami : Daddy, I don’t like the school. The headmaster beats me every day.
Swami’s Father : Why does your headmaster beat you every day without any reason? I’m sure you must be causing a lot of nuisance in school.
Swami : No, Dad. The headmaster beats all my friends in the same way.
Swami’s Father : OK. What do you want to do now? Don’t you go to school and continue your studies?
Swami : No, Dad. I’ll join some other school.

Now change the conversation into indirect speech.
Answer:
Swami’s father lovingly addressed his son and asked him why he was looking very dull. He further enquired why he had not gone to school that day. Swami replied with pain that he did not like the school and complained that the headmaster beat him everyday. Fatherdisbelievingly asked why the headmaster beat him every day without any reason. He added that he was sure Swami must be causing a lot of nuisance in school. Swami emphatically denied that and added that the headmaster beat all his friends in the same way. Father agreeingly asked what he wanted to do then. He further asked if he would not go to school and continue his studies. Swami replied that he would join some other school.

II. Noun Clause
Look at the sentences taken from the story.

1. One student said that he had an attack of a headache.
This sentence has two clauses.
a) One student said (Principal clause)
b) that he had an attack of a headache. (Subordinate clause)

The Subordinate clause is the object of the verb ‘said’. It is a noun clause. The noun clause can also appear in the subject position as can be seen in the following sentence.

2. What you say is not clear to me.
If we replace the underlined part with ‘it’ in the above sentence, the sentence structure will be complete.

Note: Sometimes the word ‘that'(conjunction) can be left out in spoken English.
e.g. He felt that punishment was not enough, (that- adjective)
Now read the following passage carefully and identify the noun clauses.
Replace the underlined words ‘that’, ‘so’, and ‘it’ with suitable noun clauses.

The headmaster entered the class furiously and said that he wanted to know the reason for the absence of some students in the class the day before. One student said that he had suffered from a severe headache. The headmaster said, “I don’t believe that”. The second said that somebody stopped him from coming to school. The headmaster said, “I don’t think so”. The third said that he too had suffered from a bad headache. On hearing that the headmaster shouted in anger. The fourth said that he had suffered from a terrible toothache. The headmaster said, “I don’t believe it”. The fifth said, “My grandmother died suddenly”. The headmaster retorted that he would ascertain it. He said, “I will come to your house tomorrow to know the fact.”
Answer:
A) Noun Clauses

1. that he wanted to know the reason for the absence of some students in the class the day before.
2. that he had suffered from a severe headache.
3. that somebody stopped him from coming to school.
4. that he too had suffered from a bad headache.
5. that he had suffered from a terrible toothache.
6. that he would ascertain it.

B) Replacing ’that1 ‘so1 and ‘it’ with suitable noun clauses.

1. “I don’t believe that.” = “I don’t believethat you have suffered from a severe headache”.
2. “I don’t thinkso” = “I don’t thinkthat somebody stopped you from coming to school”.
3. “I don’t believe it.” = “I don’t believethat you have suffered from a terrible toothache.”

III. Editing

Read the following passage and edit (correct) the underlined parts.

Swami went home and says that the headmasterbeats him severely. The parents asked that why the headmaster had beaten him. Swami said that the headmaster beats him yesterday. Swami’s father said why the headmaster has beaten him without any cause. Swami replied to his father that the headmasterbeats him every day. Swami’s mother told to Swami to attend the classes regularly.
Answer:
Swami went home and said that the headmaster had beaten him severely. The par¬ents asked why the headmaster had beaten him. Swami said that the headmaster had beaten him the day before/the previous day. Swami’s father asked Swami why the head- master had beaten him without any cause. Swami repted that the headmaster beats him every day. Swami’s mother told Swami to attend the classes regularly.

Writing

I. What do you think Swami might have thought after he had left his school? Write down his reflections in a paragraph.
Answer:
Swami controlled his tears with a great effort. He came out of the school muttering “I don’t care for your dirty school.” As he came out he must have thought thus, “What a wretched school! How cruelty has taken the shape of this headmaster? Don’t they understand our feelings? Weren’t they too children at one time? Didn’t they cause any nuisance when they were children? Yes, I did break the panes in a moment of childish behaviour. Have I not been honest in not denying the charge? Is it not enough to scold us in the whole class? He went on insulting us, beating us, and threatening us with dismissals. Is that the way to correct us? Shouldn’t they give us a chance to set ourselves right? I wish I weren’t a student of this horrible school. Thank God! At least now, I gathered enough courage to come out of the school. I can join some other school. I think any other school will definitely be better than this bloody school. Anyhow, I too should and will be more careful about my behaviour and friends. I should think of my studies and my parents too. Yes, in one way, the incident is a lesson for me. I should make utmost use of it!

II. Construction of a Narrative

Look at the concluding part of the story.
He restrained the tears that were threatening to rush out, jumped down, and, grasp¬ing his books, rushed out, muttering, ‘I don’t care for your dirty school.’
Now imagine what happens to Swami after going away from school. Write a narrative which should include dialogues, sensory perceptions etc.
You may include things such as the following.
1) Swami rushed out from the class.
2) His parents asked him what happened at school.
3) His mother looked at the scars on his shoulders.
4) His father wanted him to go to school.
5) Swami did not like to go to school.
Answer:
Swami rushed out of the school saying, “I don’t care for your dirty school.” As he walked homewards, the tears that had been controlled till then found their way out. And they came out in floods. Swami didn’t mind them. His mind was swarmed with different kinds of thoughts. His heart ached with insult. His body burned with pain. His mind was hot with a heavy flow of thoughts. He did not notice passing persons. He didn’t even take care to walk to the left of the road. In fact, his legs took him along the regularly used path. At last, somehow he reached home with school bag on his back. The face with the marks of tears, the shoulders with the prints of the cane, and the heart, most importantly, with the load of insult and suffering!

Swami’s mother was moved on seeing Swami’s pathetic state. Her eyes turned wet as she saw the red scars on Swami’s shoulders. With choked throat, she asked Swami what had happened. Even the anxious father eagerly looked forward to listening to Swami. His parents’ anxiety touched Swami. He tried to be as cool as possible. He started narrating the incident at school very briefly.

“Our headmaster beats me everyday. For one reason or the other, he scolds me, and that too in a foul way. He repeatedly warns me with dire consequences.” As he continued, it became difficult for him to control his tears. Still, he tried to suppress the sobs and paused for a while. Father took that opportunity to ask, “Why does your headmaster beat you alone everyday without any reason ?”

That question hit Swami’s heart like an arrow. Yet he restrained his feelings and said in an emotion filled voice, “Dad, he beats everyone, not me alone; that is his nature, his hobby, his practice. You can check it with any of our school students. The only difference is on the days of visits by officers or village elders.”

“Then,” interrupted mother with concern, “What shall we do ?” “I shall join some other school,” prompt was the reply from Swami lest Father should announce a different decision.

Noticing the resolute voice of Swami, even father wanted Swami’s words to prevail. Mother was too willing. She sighed in relief as there was no opposition from father. Swami started dreaming of his new school!

Study Skills

I. Read the data given in the table and answer the questions that follow.
Reasons for Dropouts among children Aged 5-14 Years – 1997-98

Source: Ministry of Human Resource Development (MHRD)
Answer the following questions:
1. What is the major reason for dropouts in rural and urban areas?
2. What percentage of female children are dropped out as their parents not being interested in their children’s studies in rural areas?
3. Which is the less significant reason for dropouts?
4. Which of the following statements are true with reference to the data given in the above table? Tick (S) the true statements.
a. The dropout-rate due to child not being interested in studies is more among the urban children than that of the rural children.
b. Participation in other economic activities is high among urban female children when compared with rural female children.
c. If we create interest in studies among the children, the literacy rate will increase in our country.
Answer:

1. The major reason for dropouts among children both in rural and urban areas is the child being not interested in studies.
2. 9.2 percent of girls are dropouts in rural areas because of lack of interest on the part of parents.
3. Working for wages/salaries is the least significant reason for dropouts.
4. a – ✗ b – ✓ c – ✓

II. Write a report on dropouts using the information given in the above table.

You may begin the report like this.
This report is based on the data provided by MHRDfor the year 1997-98for children who dropped out in the age group of 5-14 years…
Include the following.
1. The financial reasons for dropouts
2. Personal reasons for the dropouts
3. Whether the dropout rate is more among girls/boys
4. Whether the dropout rate is more among rural/urban
Answer:
This is a report on dropout rates among children aged between 5 -14. The information comes from the survey conducted by the Ministery of Human Resource Development (MHRD) for the year 1997-1998. Both the rural and urban areas (separately) were taken into consideration and the data categorises male and female children separately. Dropout rate caused by lack of interest on the part of the child tops the list with 37.2% in rural areas and 34.7% in urban areas. Working for wages is the least significant reason with just 2.5% in villages and 3.6% in towns. Dropout rate among girls is more than that of the boys by 10% in rural areas and by about 6% in urban areas. Economic, social, personal, and other reasons contribute for this undesirable state of affairs.

Listening

Practise listening carefully. Then you will be able to speak.
Listen to the 2 speeches and answer the questions that follow.

Speeches!
Speech 1
Good evening to all the people present here today. It’s a pleasure to start this occasion by welcoming everybody. I welcome our headmaster to preside over the function. Now I would like to invite the honourable chief guest, our M.L.A to come onto the dais. I welcome my teachers, my fellow – friends, and schoolmates. Today we all have gathered here for the Annual Day celebration. I welcome you all and I hope you enjoy the programme.
Now I request our headmaster to start the function.
Thank you.

Speech 2 :
I have a great pleasure in welcoming our principal to preside over the Children’s Day programme that we have today. It is a great privilege for me to invite our chief guest, the D.E.O. of our district, who has kindly consented to be the chief guest for the day, I also welcome other distinguished guests, who also have consented to be with us today despite their tight schedule. I cordially welcome my colleagues and non-teaching staff too. In today’s celebration our main heroes are our students. So I welcome them and their parents too along with all others.

I hope you enjoy every aspect of this event and request the president to begin the proceedings.
Thank you.
Now answer the following questions:
1. Who is the speaker of speech 1?
Answer:
A student is the speaker of speech 1.

2. What is the occasion mentioned in the first speech?
Answer:
The occasion is Annual Day celebration. (1st speech)

3. What is the occasion mentioned in the second speech?
Answer:
The occasion is Children’s Day programme. (2nd speech)

4. Who is the chief guest mentioned in the second speech?
Answer:
The D.E.O of the district is the chief guest. (2nd speech)

5. Who is addressing the gathering in the second speech?
Answer:
A teacher is addressing the gathering. (2nd speech)

Oral Activity

Imagine that you are the School Pupils’ Leader (SPL). Compere on the Republic Day celebrations in your school.
You may include the following in your speech :
1) Welcome address
2) Inviting the guests onto the dais
3) Importance of the occasion
4) Request to continue the proceedings
Answer:
A brilliant morning to every soul gracing the grand occasion of our Republic Day celebrations. A noble national festival and our patriotic feelings have brought us all together here. On this memorable event, I extend my warm welcome to you all and the greetings of the day too. Now I invite our honourable headmaster to come over to the dais to preside over the proceedings. I feel it my privilege and pleasure to invite our distinguished guest of the day, our beloved collector on to the dais. Let us all extend our thunderous applause as our collector comes to grace the occasion. It is my pleasure to invite on to the dais our dear teachers who have won our hearts with their excellent teaching skills and touching attitude.

Now my dear fellow students, parents, and guests let us cherish each moment of the programmes it gradually unfolds, majestically moves ahead, and colourfully culminates. May I now request the president of the programme to lead the proceedings.
Thank you one and all.

Swami is Expelled from School Summary in English

Rasipuram Krishnaswamy Narayan’s novel Swami and Friends is an interesting story of Swami. The present part deals with one incident in Swami’s school life. One day Swami and his friends did not attend classes. Moreover, they broke the window panes of the Headmaster’s room. The peon saw that act. The following day the headmaster entered Swami’s class with a cane and a furious face. He started scolding them. He also demanded every one to explain to him why they had been absent the previous day. When they cooked up some stories, he asked for proof. He used the cane, asked them to stand up on their benches, announced suspension, etc. When it was the turn of Swami, he had no stories to tell the headmaster. His silence added insult to the headmaster’s injury. He went on beating Swami on the shoulders, the back with his cane. Swami’s complaints of pain made the headmaster more violent. Not able to bear the torture any more, Swami left the class

Swami is Expelled from School Glossary

ominous (adj) : indicating the happening of bad

but (preposition) (here) : except

eminence (n) : fame, respect

loafer(n) : a person who wastes time without working

ascertain (v) : find to be true or not

pause(n) : a short stop ; gap

knuckles (n-plural-the first ‘k’ silent : joints in fingers
(Note : The ‘k’ at the beginning of a word followed by ‘n’ is silent, e.g. : know, knife, knight)

resolutely (adv) : with a strong decision

gazing (v-ing) : looking

intently (adv) : with all attention

acute (adj) : intense

rap (n) : a sharp hit

gaol(n) : jail, prison

discreet (adj) : careful; tactful

stammered (v-past tense) : spoke with difficulty ; repeating sounds

thrash (v) : hit, beat

sinister (adj) : evil, dangerous

grunted (v-past tense) : made short, low sounds in the throat to show Irritation

ventilators (n-plural) : window like arrangements close to the ceiling

zest (n) : enthusiasm

whack (n) : a hit

brigand (n) : a member of a group of criminals

defiant (adj) : refusing to obey

deny (v) : to refuse to admit

charge (n) : accusation

ejaculated (v-past tense) : shouted suddenly

staring (v+ing) : looking

idiot (n) : a stupid person ; a fool

desperation (n) : hopeless condition

restrained (v – past tense) : controlled

muttering (v + ing) : saying in a low voice

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 9 Statistics InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 9th Lesson Statistics InText Questions

Activity

Divide the students of your class into four groups. Allot each group the work of collecting one of the following kinds of data: [Page No. 195]
Solution:
i) Weights of all the students in your class
ii) Number of siblings that each student have.
iii) Day wise number of absentees in your class during last month.
iv) The distance between the school and home of every student of your class.

Do This

Which of the following are primary and secondary data ? (Page No. 195)
i) Collection of the data about enroll¬ment of students in your school for a period from 2001 to 2010.
Solution:
Secondary data.

ii) Height of students in your class re¬corded by physical education teacher.
Solution:
Primary data.

Activity

Make frequency distribution table of the initial letters of that denotes surnames of your classmates and answer the following questions.
i) Which initial letter occured mostly among your classmates?
ii) How many students names start with’ the alphabat T?
iii) Which letter occured least as an initial among your classmates? [Page No. 197]

Think, Discuss anil Write

Question 1.
Give 3 situations, where mean, median and mode are separately appropriate and counted. [Page No. 202]
Solution:
Mean :
a) Rice required for a certain number of students for a given period.
b) To compare the marks of students of a class.
c) To calculate the daily income of a shop during a month.

Median :
a) To study the salaries of staff of an institution.
b) To compare heights of boys and girls.

Mode :
a) To find the size of the shoe with heighest sale.

Try These

Question 1.
Find the median of the scores 75, 21, 56, 36, 81, 05, 42. [Page No. 207]
Solution:
Arranging the data in ascending order
05, 21, 36, 42, 56, 75, 81
Number of terms in the data = 7 – odd
∴ \(\left(\frac{n+1}{2}\right)^{t h}\) term = \(\frac{7+1}{2}=\frac{8}{2}\) = 4th term is the median = 42.

Question 2.
Median of a data arranged in ascending order 7, 10, 15, x, y, 27, 30 is 17 and when one more observation 50 is added to the data the median has become 18. Find x and y. [Page No. 207]
Solution:
The given data in ascending order is 7, 10, 15, x, y, 27, 30
Median = \(\left(\frac{n+1}{2}\right)^{t h}\) term = \(\left(\frac{7+1}{2}\right)\) = 4th term = x
∴ x = 17 by problem
When 50 is added, the data becomes 7, 10, 15, 17, y, 27, 30, 50
Median = \(\frac{\left(\left(\frac{n}{2}\right)+\left(\frac{n}{2}+1\right)\right)}{2}\) terms
18 = \(\frac{17+y}{2}\) (by problem)
17 + y = 36
y = 36 – 17 = 19

Question 3.
Find the median marks in the data.[Page No. 208]

Solution:

Median = \(\left(\frac{n+1}{2}\right)^{t h}\) term as N = 29 is odd
∴ \(\frac{29+1}{2}\) = 15th term.
15th observation is 15. (from the table)

Question 4.
In finding the median, the given data must be in order ? Why ? [Page No. 208]
Solution:
In finding the median, the observations must be in order. The data is to be arranged either in ascending/descending order to divide the data into two equal groups.

Think and Discuss

Question 1.
Classify your classmates according to their heights and find the mode of it.
(Page No. 208)
Solution:
Student to find the mode of the classmates according their heights.

Question 2.
If shopkeeper has to place an order for shoes, which number shoes should he order more ? [Page No. 208]
Solution:
Number 7 as it has highest sales.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 8 Quadrilaterals InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals InText Questions

Try This

Question
Extend AB to E. Find ∠CBE. What do you notice? What kind of angles are ∠ABC and ∠CBE?
Solution:
Given that □ABCD is a parallelogram and∠A = 40°
∴ ∠ABC = 180°-40° = 140°
∠CBE = 40° ( ∵ ∠A and, ∠CBE are corresponding angles) And ∠CBE and ∠ABC are linear pair of angles.

Do This

Question
Cut out a parallelogram from a sheet of paper again and cut along one of its diagonal. What kind of shapes you obtain ? What can you say about these triangles 7 [Page No. 179]
Solution:
We get two triangles.

The two triangles are congruent to each other.

Think, Discuss and Write

Question 1.
Show that the diagonals of a square are equal and right bisectors of each other. (Page No. 185)
Solution:

Let □ABCD be a square.
Thus AB = BC = CD = DA
In ΔABC and ΔBAD
AB = AB (common base)
∠B =∠A (each 90°)
BC = AD (equal sides)
∴ ΔABC = ΔBAD (SAS congruence)
⇒ AC = BD (CPCT)
Also in ΔAOB and ΔCOD
∠OAB = ∠OCD [∵ alt. int. angles]
∠OBA = ∠ODC [∵ alt. int. angles]
AB = DC (sides of a square)
∴ ΔAOB = ΔCOD (ASA congruence)
Thus AO = OC (CPCT) ⇒ O is midpoint of AC
Also BO = OD (CPCT) ⇒ O is midpoint of BD
∴ O is midpoint of both AC and BD.
∴ Diagonals bisect each other.
In ΔAOB and ΔCOB
AB = BC (given)
OB = OB (common)
AO = OC (proved)
∴ ΔAOB ≅ ΔCOB (SSS congruence)
∠AOB = ∠COB (CPCT)
But ∠AOB + ∠COB = 180° (∵ linear pair of angles)
∴ ∠AOB = ∠COB = 180°/2 = 90°
Also ∠AOB = ∠COD (∵ vertically opposite angles)
∠BOC = ∠AOD (∵ vertically opposite angles)
∴ AC ⊥ BD
(i.e.,) In a square diagonals bisect at right angles.

Question 2.
Show that the diagonals of a rhombus divide it into four congruent triangles. (Page No. 185)
Solution:

□ABCD is a rhombus
Let AC and BD meet at ‘O’
In ΔAOB and ΔCOD
∠OAB = ∠ODC (alt.int.angles)
AB = CD (def. of rhombus)
∠OBA = ∠ODC (alt. mt, angles)
∴ ΔAOB ≅ΔCOD …………(1)
(ASA congruence)
Thus AO = OC (CPCT
Also ΔAOD ≅ ΔCOD …………..(2)
[ ∵AO = OC; AD = CD; OD = OD SSS congruence]
Similarly we can prove
ΔAOD ≅ ΔCOB …………..(3)
From (1), (2) and (3) we have
ΔAOB = ΔBOC = ΔCOD = ΔAOD
∴ Diagonals of a rhombus divide it into four congruent triangles.

Try This

Question
Draw a triangle ABC and mark the mid points E and F of
two sides of triangle. \(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{A C}}\) respectively. Join the point E and F as shown in the figure. Measure EF and the third side BC of triangle. Also measure ∠AEF and ∠ABC.
We find ∠AEF = ∠ABC and \(\overline{\mathrm{EF}}=\frac{1}{2} \overline{\mathrm{BC}}\)

As these are corresponding angles made by the transversal AB with lines EF and BC, we say EF//BC.
Repeat this activity with some more triangles. (Page No. 188)

Solution:

P, Q are mid points of XY and XZ
PQ // YZ
PQ = \(\frac{1}{2}\)YZ

X, Y are mid points of PQ and PR
XY // QR
XY = \(\frac{1}{2}\)QR

A, B are midpoints of DE and DF
AB // EF
AB = \(\frac{1}{2}\)EF
( ∵ In all cases the pairs of respective corresponding angles are equal.)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 7 Triangles InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles InText Questions

Do This

Question 1.
There are some statements given below. Write whether they are true or false. [Page No. 150]
Solution:
i) Two circles are always congruent. – False
ii) Two line segments of same length are always congruent. – True
iii) Two right angle triangles are some times congruent. – True
iv) Two equilateral triangles with their sides equal are always congruent. – True

Question 2.
Which minimum measurements do you require to check if the given figures are congruent? [Page No. 150]

i) Two rectangles.
Solution:
Length and breadth are required.

ii) Two rhombuses.
Solution:
One side and one interior angle are required.

Do These

Question 1.
State whether the following triangles are congruent or not. Give reasons for your answer. [Page No. 153]

Solution:
i) ΔABC ≅ ΔDEF
∵ ∠B = ∠E (∵ Angle sum property Z E = 180° – (70° + 60°) = 50°)
BC = EF
∠C = ∠F
∴ By SAS congruence
ΔABC ≅ ΔDEF

ii) In ΔMNL and ΔTSR
MN = ST
NL = TR
∠ M = ∠ T (∵ ΔMNL is flipped to get ΔTSR)
∴ ΔMNL ≅ ΔTSR

Question 2.
In the given figure, the point P bisects AB and DC. Prove that ΔAPC ≅ ΔBPD. [Page No. 153]

Solution:
Given that P bisects AB and DC.
Now in ΔAPC and ΔBPD
AP = BP (∵ P bisects AB)
CP = DP (∵ P bisects CD)
∠ APC = ∠ BPD
∴ ΔAPC ≅ ΔRPP (∵ SAS congruence)

Activity

1. On a tracing paper draw a line segment BC of length 6 cm.
2. From vertices B and C draw rays with angle 600 each. Name the point A where they meet
3. Fold the paper so that B and C fit precisely on top of each other. What do you observe? Is AB = AC?
[Page No. 160]

Do This

Question 1.
In the figure given below ΔABC and ΔDBC are two triangles such that \(\overline{\mathbf{A B}}=\overline{\mathbf{B D}}\) and \(\overline{\mathbf{A C}}=\overline{\mathbf{C D}}\) . Show that ΔABC ≅ ΔDBC [Page No. 164]

Solution:
Given that \(\overline{\mathbf{A B}}=\overline{\mathbf{B D}}\) and \(\overline{\mathbf{A B}}=\overline{\mathbf{B D}}\)

In ΔABC and ΔDBC
AB = BD (∵ given)
AC = DC (∵ given)
BC = BC (common side)
∴ ΔABC ≅ ΔDBC
( ∵ by SSS congruence)

Activity

Question 1.
Construct a right angled triangle with hypotenuse 5 cm. and one side 3 cm. long. How many different triangles can be constructed? Compare your triangle with those of the other members of your class. Are the triangles congruent? Cut them out and place one triangle over the other with equal side placed on each other. Turn the triangle if necessary what do you observe? You will find that two right triangles are congruent, if side and hypotenuse of one triangle are respectively equal to the corresponding side and hypotenous of other triangle.
[Page No. 165]

Activity

Question 1.
Draw a triangle ABC mark a point A’ on CA produced (new position of it)
So A’C > AC (Comparing the lengths) Join A to B and complete the triangle ABC.

What can you say about ∠A’BC and ∠ABC ?
Compare them. What do you observe?

Clearly, ∠A’BC > ∠ABC Continue to mark more points on CA (extended) and draw the triangles with the side BC and the points marked. You will observe that as the length of the side AC is increases (by taking different positions of A), the angle opposite to it, that is ∠B also increases.
[Page No. 169]

Question 2.
Construct a scalene triangle ABC (that is a triangle in which all sides are of different lengths). Measure the lengths of the sides.

Now, measure the angles. What do you observe?
In ΔABC figure, BC is the longest side and AC is the shortest side. ‘
Also, ∠A is the largest and ∠B is the smallest.
Measure angles and sides of each of the above triangles, what is the rela tion between a side and its opposite angle when compared with another pair?

Activity

Question
Draw a line segment AB. With A as centre and some radius, draw an arc and mark different points say P, Q, R, S, T on it.
Solution:

Join each of these points with A as well as with B (see figure). Observe that as we move from P to T, ∠A is becoming larger and larger. What is happening to the length of the side opposite to it? Observe that the length of the side is also increasing; that is ∠TAB > ∠SAB > ∠RAB > ∠QAB > ∠PAB
and TB > SB > RB > QB > PB.
Now, draw any triangle with all angles unequal to each other. Measure the lengths of the sides (see figure).

Observe that the side opposite to the largest angle is the longest. In figure, ZB is the largest angle and AC is the longest side
Repeat this activity for some more triangles and we see that the converse of the above Theorem is also true.

Measure angles and sides of each triangle given below. What relation you can visualize for a side and its opposite angle in each triangle.

In this way, we arrive at the following theorem. [Page No. 170]

Do This

Question
Draw a triangle ABC and measure its sides. Find the sum of the sides AB + BC, BC + AC; and AC + AB, compare it with the length of the third side. What do you observe ?
[Page No. 171]
Solution:

AB + BC = 4 + 3 = 7
= 7 > 4 = AC
BC + CA > AB;
3 + 4 > 4
CA + AB > BC;
4 + 4 > 3

DE + EF > DF
EF + DF > DE
FD + DE > EF
∴ Sum of any two sides of a triangle is greater than the third side.

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 2C V.V.S. Laxman, Very Very Special Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 2C V.V.S. Laxman, Very Very Special

9th Class English Chapter 2C V.V.S. Laxman, Very Very Special Textbook Questions and Answers

I. Answer the following questions.

Question 1.
What is Laxman’s philosophy of life as per the interview you have read?
Answer:
Laxman’s philosophy of life is to work hard continuously and treat success and failure equally.

Question 2.
What role did Laxman play in making India, No. 1 test cricket team?
Answer:
Laxman says that India hadn’t become No.l all of a sudden. It was a slow and long drawn process. He was frank in saying that he did not play in the World Cup that saw India as the No. 1 team. But he was proud to be the member of the world No.l team.

On the basis of your reading of V.V.S. Laxman’s interview with Sportstar magazine complete the following table.

Events/Incidents in his life	Your responses
1. Home ground
2. Teams he represented
3. Levels he played at
4. People who influenced him
5. Things he likes /hobbies
6. Memorable events
7. His message

Answer:

Events/Incidents in his life	Your responses
1. Home ground	Hyderabad Cricket Association – Uppal Stadium
2. Teams he represented	Under 16; Under 19; Ranji Trophy – Hyderabad team; National team, One Day Internationals; T20s – IPL
3. Levels he played at	Under 16; Under 19; Senior Level
4. People who influenced him	Parents; Specially father Dr. V. Shantaram
5. Things he likes /hobbies	Spending with members of family; reading biographies of successful individuals; listening to music.
6. Memorable events	281 against Australia in Kolkata in 2001
7. His message	Work hard continuously; treat success and failure equally

Writing

Write persona! views and reflections on V.V.S. Laxman in a paragraph of about 75-100 words.
Discuss the following questions in groups before writing the paragraph individually.
1) What is the main idea that you wish to project?
2) What are the supporting ideas that you think of?
3) How do you organize your ideas?
4) How do you link your thoughts?
5) How do you conclude?
Answer:
V.V.S. Laxman is both an excellent cricketer and a wonderful individual. As a cricketer, he reached dizzy heights and won for him a special title ‘Very Very Special.’ But as a human being, he exhibits very noble qualities. He loves his parents, wife, and children intensely. He is thankful to all those who helped him. He is frank in admitting his weak points. He is honest in declaring his sources of inspiration. He is free to announce his hobbies, food habits, and future plans. His philosophy to treat success and failure equally is highly admirable. On the whole, V.V.S. Laxman is a very, very special model to everyone.

Project Work

Collect the information from newspapers, magazines, periodicals, and books about two famous Indian sportswomen and prepare their profiles.

Fill the details of the following information and you may use them as tips for profile writing and speaking.

Talk about one profile in the class.

Profile – 1: (P.T.Usha)
Name	Pilavullakandi Thekkeparambil Usha
Date of Birth	June 27, 1964
Height	5.7″ (170 cm)
State/Team she represented	National Team – India
Sports/Game she is associated with	Running (Track & Field)
Debut (first entry)	At the age of 16, in 1980 in the Moscow Olympics
Best in the Career	At the Asian Meet in Jakarta in 1985, Usha established herself as the Asian sprint queen with five gold medals (in the 100 meters, 200m, 400m, 400m hurdles, and the 4 x 400 m relay) besides a bronze in the 100m relay,
Hobbies	Reading books and listening to music

Answer:
Awards/Medals Received

1. Recipient of Arjuna Award, 1984.
2. Adjudged as the greatest women athlete, in 1985 Jakarta Asian Athletic Meet.
3. Padma Sree in 1984.
4. Best Athlete of the year Award from India Government in 1984,/85,/B6,/B7, and 89.
5. In 1986 Seoul Asian Games, won the Adidas Golden Shoe Award for the best athlete by the Asian Amateur Athletics Association, Seoul Asian Games, 1986.
6. Asian Amateur Athletics Association, Seoul Asian Games, 1986.
7. 33 medals including 13 gold medals in Asian Games and Asian Championships.
8. Won a total of 102 medals at National and International meets during her career.
9. Won 1 gold and 2 silver at the 1999 SAF Games held at Kathmandu.
10. Thirty International Awards, for her excellence in Athletics.
11. Kerala Sports Journalists Award for the year 1999.

Profile with more details

P.T. Usha was born, as the daughter of Paithal and Lakshmi, at Payyoli, a small village in Kozhikode, on June 27, 1964. Her full name is Pilavullakandi Thekkeparambil Usha. She was the queen of Indian track and field for two decades. P.T. Usha has been associated with Indian athletics since 1979. Usha made her international debut at the Moscow Olympics in 1980 but she shone into the limelight in the 1982 Asian Games in New Delhi, winning the silver in the 100 m and 200 m event. At the Asian Meet in Jakarta in 1985, Usha established herself as the Asian sprint queen with five gold medals (in the 100meters, 200m, 400m, 400m hurdles, and the 4 x 400m relay) besides a bronze in the 100m relay.

This magnificent performance was followed by an equally brilliant spell a year later at the Asian Games at Seoul where Usha notched up four golds and a silver medal.

The finest moment in Usha’s career and also perhaps the saddest however came in a single race at the 1984 Olympics in Los Angeles. In the 400m hurdles, Usha missed winning the bronze by just 1/100th of a second. She recorded her best time of 55.42secs in that race — still an Indian national record — but lost the medal in a photo-finish. Usha said that she cried after the event because “It was difficult to believe that I had missed an Olympic medal by a whisker.”

In 1976 the Kerala State Government started a Sports School for women, and Usha was chosen to represent her district, at a cost of Rs. 250 per month paid by the state. In 1979 she participated in the National School Games, where she was noticed by O. M. Nambiar, who coached her through most of the rest of her career. India Today describes the athletic situation in 1979 as a time when ‘athletics was very much a male sport and track-suited women a rarity’.

P.T Usha started a School of Athletics to impart training to girl children from all over the country. The School, located at Koyilandi near Kozhikode in Kerala, recruits children in the 10-12 age group for its training. She likes reading books and listening to music.

Awards and Medals Received

1. Recipient of Arjuna Award, 1984.
2. Adjudged as the greatest women athlete, in 1985 Jakarta Asian Athletic Meet Padma Sree in 1984.
3. Best Athlete of the year Award from India Government in 1984,/85,/86,/87, and 89.
4. In 1986 Seoul Asian Games, won the Adidas Golden Shoe Award for the best athlete by the Asian Amateur Athletics Association, Seoul Asian Games, 1986.
5. 33 medals including 13 golds in Asian Games and Asian Championships.
6. Won a total of 102 medals at National and International meets during her career.
7. Won 1 gold and 2 silver at the 1999 SAF Games held at Kathmandu.
8. Thirty International Awards, for her .excellence in Athletics.
9. In recognition of her achievements, a road at Payyoli, her home town, is named after her.
10. The Kerala Government has also set up a “PT Usha Sports Council” at Central Stadium, Thiruvananthapuram.
11. Kerala Sports Journalists Award for the year 1999.

Profile – 2

Profile – 2: (Koneru Humpy)
Name	Koneru Humpy
Date of Birth	31 March, 1987
Height	5.6″ (165 cm)
State/Team she represented	National – India
Sports/Game she is associated with	Chess
Debut (first entry)	At the age of 8, in 1995 in Indian Under 8 Championship
Best in the Career	The second woman in the world with 2606 points in FIDE rating in July 2009
Hobbies	Reading books and spending with family

Awards and Medals Received

1. Asia’s youngest International Woman Master, 1999.
2. World under-14 championship, 2001, Castellan, Spain.
3. India’s youngest Woman Grand Master, 2001.
4. World Junior Championship, 2001, Athens.
5. World’s youngest Women Grandmaster to achieve full Grand Master status.
6. Arjuna Award in the year 2003.
7. In 2007, she was awarded with the prestigious Padma Shri award.
8. Humpy was also conferred upon the Raja Lakshmi Award in the year 2008 by Raja Lakshmi Foundation of Chennai.
9. become the second-highest ranked female player in history with more than 2600 points in FIDE rating in July 2009.

Profile with more details

Koneru Humpy is a popular femaie Indian chess player, and feasibly the best woman at the chess board. She is the world second ranking among the Female Chess PIaryers,’stayed behind only by Judit Polgar, who is world number one Female Chess Player.

She was born on 31st March, 1987 at Gudivada, Andhra Pradesh. Her father is Ashok Koneru worked as a lecturer in Chemistry and he was a quite well chess player. In 1985 he won the South India Open Championship. Humpy fell in love with the game of chess when she was just 5 years of old. In fact, in orderto guide her properly and to make sure she gets the best attention for improving her skills as a chess player her father introduced her to the game at quite an early age. The little Humpy showed her outstanding performance in chess and she won the Under 8 National Chess Championship in 1995.

After proved her brilliant performance at the National level, Humpy entered the international chess circle. She clinched the World Chess titles in the Under 10, Under 12 and Under 14 age groups, later, in 1990 Humpy holds an International Master title when she was 12 years. After, she gained her 3rd Grand Master norm in the Elekes Memorial Grand Master Tournament held at Budapest, Hungary. Koneru has set up a world record by getting the International Grand Master title at the age of 15 years old. She broke Judit Polgar record to achieve the feat, and she became the youngest woman ever to have got the coveted title. Further Hou Yifan broke Humpy’s record by taking the title, when she was 14 years. Humpy has been the First Indian Woman to have achieved an International Grand Master title in the chess game.

She likes reading books and spending with her family

For showing advanced talent as a chess player and making the nation proud at many times at the International level, Humpy has been honoured with a number of awards and recognitions.

Awards & Honours

1. Asia’s youngest International Woman Master, 1999.
2. World under-14 championship, 2001, Castellan, Spain.
3. India’s youngest Woman Grand Master, 2001.
4. World Junior Championship, 2001, Athens.
5. World’s youngest Women Grandmaster to achieve full Grand Master status.
6. Arjuna Award in the year 2003.
7. In 2007, she was awarded with the prestigious Padma Shri award.
8. Humpy was also conferred upon the Raja Lakshmi Award in the year 2008 by Raja Lakshmi Foundation of Chennai.
9. At Doha Asian Games 2006, Koneru Humpy bagged two Gold Medals in the Individual as well as Team event of Chess.
10. In 2007, she won the International Open Chess Tournament 2007 held at Kaupthing, Luxembourg.
11. Humpy scored a FIDE Elo rating of 2606 points
12. Humpy has broken the world record set by Susan Polgar who had a rating of 2577 points while she was at the World No. 2 spot,

V.V.S. Laxman, Very Very Special Summary in English

Vangipurapu Venkata Sai Laxman recently retired from cricket. Sportstar interviewed this sports star on that occasion. Laxman gives us lots of details. He says his parents Dr. V Shantaram and Dr. V. Satyabhama were his inspiration and influence. He adds that the Hyderabad Cricket Association encouraged him even as a boy. He acknowledges his uncle Baba Mohan’s help and his coaches’ guidance.

He cherishes his 281 against Australia in Kolkata in 2001. He remembers his embarrassment when he collided against Sourav in an Oval ODI. He fondly recollects his association with his captains Sourav, Sachin and Dhoni. He admires John Wright’s role as a coach. He repeatedly appreciates his wonderful wife and lovely kids. He is proud of his role in making India No. 1 in cricket. He is happy to put an end to his 16 year long loving cricket career. His message to the young is to work hard and to treat success and failure with equal ease. He gratefully declares that cricket has taught him character. He plans to start a school and an academy. He is confident of his success in his future ventures too.

V.V.S. Laxman, Very Very Special Glossary

fabulous (adj) : fantastic, excellent

leap (n) (here) : progress

on the verge of : very close to

culminated (v-past tense) : resulted in; ended

integral (adj) : essential, main

crucial (adj) : valuable, important

reckoning (n) : consideration

transformation (n) : change; new form

traits (n) (plural) : qualities

amazingly (adv) : surprisingly impressive

collided (v-past tense) : dashed against

disgusted (v-past tense) : disappointed

cuisine (n) : dish

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables InText Questions

Try This

Question 1.
Express the following linear equations in the form of ax + by + c = 0 and indicate the value of a, b, c in each case. [Page No. 128]
i) 3x + 2y = 9
Solution:
3x + 2y = 9
⇒ 3x + 2y – 9 = 0
Here a = 3, b = 2, c = -9

ii) – 2x + 3y = 6
Solution:
– 2x + 3y = 6
⇒ – 2x + 3y – 6 = 0
Here a = -2, b = 3, c = -6

iii) 9x – 5y = 10
Solution:
9x- 5y = 10
⇒ 9x- 5y – 10 = 0
Here a = 9, b = -5, c = -10

iv) \(\frac{x}{2}-\frac{y}{3}-5=0\)
⇒ \(\frac{3 x-2 y-30}{6}=0\)
⇒ 3x – 2y – 30 = 0
Here a = 3, b = – 2 and c = – 30

v) 2x = y
Solution:
2x = y
⇒ 2x – y = 0
Here a = 2, b = – 1 and c = 0

Try These

Take a graph paper, plot the point (2, 4), and draw a line passing through it. Now answer the following questions. [Page No. 135]

Question 1.
Can you draw another line that passes through the point (2, 4) ?
Solution:
Yes, We can draw another line passing through (2, 4).

Question 2.
How many such lines can be drawn ?
Solution:
Infinitely many lines can be drawn.

Question 3.
How many linear equations in two variables exist for which (2, 4) is a solution ? Solution:
Infinitely many linear equations in two variables exist.
Graph is given in next page.

Do This

Question 1.
i) Draw the graph of following equations. [Page No. 145]
a) x = 2 b) x = -2 c) x = 4 d) x = -4

ii) Are the graphs of all these equations parallel to Y-axis? .
Solution:
Yes; all these lines are parallel to Y-axis.

iii) Find the distance between the graph and the Y-axis in each case.
Solution:

Graph	DIstance	Right side/left side of the origin
a) x = 2	2 units	Right side
b) x = -2	2 units	Left side
c) x = 4	4 units	Right side
d) x = -4	4 units	Left side

Question 2.
i) Draw the graph of the following equations. [Page No. 145]
a) y = 2 b) y = -2 c) y = 3 d) y = – 3
Solution:

ii) Are all these parallel to the X-axis?
Solution:
Yes, these are all parallel to X-axis.

iii) Find the distance between the graph and the X-axis in each case.

Graph	DIstance	side of the origin above/below
a) y = 2	2 units	above origin
b) y = -2	2 units	below origin
c) y = 3	3 units	above origin
d) y = -3	3 units	below origin

Try These

Question
Find the equation of the Y – axis. [Page No. 146]
Solution:
All points on Y – axis have their x – co-ordinates as zero.
∴ Equation of Y- axis is x = O.

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 2B What is a Player? Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 2B What is a Player?

9th Class English Chapter 2B What is a Player? Textbook Questions and Answers

I. Answer the following questions.

Question 1.
What are the qualities that a true player should have?
Answer:
A true player accepts defeat with ease. He learns from his failures. He loves to share and enjoys the rival’s victory too. He knows his abilities. He continuously works for progress. He practises in all seasons and at all times. He never complains. He is not worried about past failures. He thinks of the present and plans for the future.

Question 2.
What, according to the poet, is an unacceptable crime of a player?
Answer:
Complaining about play time is an unacceptable crime.

Question 3.
Which game do you like the most? What are the qualities of a player stated in the poem? Which of them do you have?
Answer:
I like volleyball the most. The poem lists the qualities of a player. They are accepting defeat, learningfrom failures, continuous practice, knowing one’s own limitations, sharing, not complaining, not worrying about past defeats, focussing on the present, and contributing to the game. I have some of them. They are : accepting defeat, regular practice, learningfrom failures, not complaining.

Question 4.
The poet talks about certain Do’s and Don’ts for a true player. List them out in the following table. One is done for you.

Do’s	Don ts
regular practice	give up at the sight of defeat

Answer:

Do’s	Don ts
1. regular practice	give up at the sight of defeat
2. learn from mistakes	give up at the sound of the buzzer
3. have sensible mind	whine/complain
4. contribute to the game	settle for less
5. admit to be fine even when hurt	chicken out

What is a Player? Summary in English

Jessica Taylor’s poem, “What is a Player” lists the qualities a good player possesses. An ideal player continues despite defeat and learns from failures. A good player finds pleasure in sharing and enjoys the rival’s victory too. A true sportsman never complains about his chances, for he knows it is a crime. A model player practises regularly and keeps in mind that success is not always guaranteed. The real player always aims high, works for continuous advancement, and practises day and night; in rain and shine and when healthy or not. A player is always sensible and never worries about past defeats and thinks of the present and plans for the future. A true player doesn’t do something in fear and wants to contribute to the game as long as he is there.

What is a Player? Glossary

give up (phr.v) : quit, leave

buzzer(n) : bell

whine (v) : complain

hurt (adj) : injured

sensible (adj) : reasonable, logical, sane

pondering (v + ing) : thinking

contemplating (v) : thinking; planning

chicken out (phr. v) : to decide not to do something because of fear

counts (v) : be important

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 5th Lesson Co-Ordinate Geometry InText Questions

Do This

Question
Describe the seating position of any five students in your classroom. [Page No. 108]
Solution:
Oral answer – depend on the classroom.

Activity (Ring Game)

Question
Have you seen ‘Ring game’ in exhibitions ? We throw rings on the objects arranged in rows and columns. Observe the following picture. [Page No. 108]
Complete the following table.
Solution:

Question
Is the object 3rd column and 4th row is same as 4th column 3rd row ?
Solution:
No

Do This

Question
Among the points given below, some of the points lie on X – axis. Identify them.
i) (0, 5) ii) (0, 0) iii) (3, 0) iv) (- 5, 0) v) (- 2, – 3) vi) (- 6,0) vii)(0,6) viii)(0, a) ix) (b, 0) [Page No. 114]
Solution:
The points (ii) (0, 0), (iii) (3, 0) (iv) (- 5, 0) (vi) (- 6, 0) and (ix) (b, 0) lie on X – axis.

Try These

Question 1.
Which axis the points such as (0, x), (0, y), (0, 2) and (0, – 5) lie on ? Why ? [Page No. 114]
Solution:
All the above points lie on Y – axis, since the X – coordinate of all these points is zero.

Question 2.
What is the general form of the points which lie on X – axis. [Page No. 114]
Solution:
The general form of points lying on X – axis is (x, 0).

Do This

Question
Plot the following points on a Cartesian plane.
1. B (- 2, 3) 2. L (5, – 8) 3. U (6, 4) 4. E (-3, – 3) [Page No. 120]
Solution:

Do This

Question
i) Write the co-ordinates of the points A, B, C, D and E. [Page No. 121]
ii) Write the co-ordinates of F, G, H, I and J.

Solution”
i) A(2, 9) ; B(5, 9); C(2, 6) ; D(5, 3) ; E(2, 3)
ii) F(- 6, – 2) ; G(- 4, – 5) ; H(- 3, – 7) ; I(- 9, – 7) ; J(- 8, – 5)

Activity

Question
Study the positions of different cities like Hyderabad, New Delhi, Chennai and Vishakapatnam with respect to longitudes and latitudes on a globe. [Page No. 123]
Solution:
Student Activity

Creative Activity

Question
Take a graph sheet and plot the following pairs of points on the axes and join them with line segments. [Page No. 123]
(1, 0) (0, 9) ; (2, 0) (0, 8) ; (3, 0) (0, 7) ; (4, 0) (0, 6) ;
(5, 0) (0, 5) ; (6, 0) (0, 4) ; (7, 0) (0, 3) ; (8, 0) (0, 2) ; (9, 0) (0, 1).
Try to complete the picture by using above points. What did you observe ?
Solution:
Student Activity

AP Board 9th Class Maths Solutions

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 2A True Height Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 2A True Height

9th Class English Chapter 2A True Height Textbook Questions and Answers

Read the following quotation and answer the questions that follow.

Question 1.
What, according to you, does this quotation express?
Answer:
The quotation tells us to dream first, think about it next and work on it finally.

Question 2.
Is it essential to dream high? How can anyone make one’s dream come true?
Answer:
Yes, it is good to dream high. Our dear Teacher Dr. APJ Abdul Kalam rightly says, “small aim is a big crime”.

One can make one’s dream come true with careful planning and continuous, consistent work with confidence, competence and with dedication, and discipline.

Question 3.
What is your dream in life?
Answer:
My dream in life is to become a man-making teacher and work for quality human resource development.

Question 4.
Have you ever heard of a child with special needs becoming a champion?
Answer:
Yes, I have heard of a child with special needs becoming a champion. A sixteen year old child with multiple complaints completed the cycling race much ahead of hundreds of healthy and older participants

Comprehension

I. Answer the following questions.

Question 1.
What was Michael Stone’s dream?
Answer:
Michael Stone’s dream was to become a champion in pole-vaulting.

Question 2.
What impact did the mother’s stories have on Michael?
Answer:
Michael’s mother, Mildred Stone was a good story teller. When Michael was growing up, she told him a number of stories about flying. They were full of details and colour. They inspired Michael to dream high. In his dreams, he would outrun motorcars and fly like an eagle.

Question 3.
How did Michal prepare himself to become an Olympic champion?
Answer:
Even as a young boy of 14, Michael started working hard. He began a careful, disciplined weightlifting programme. On alternate days, he practised running. His father supervised his preparation. Michael always strived for perfection. Thus his preparation to become an Olympic champion was planned, complete and perfect.

Question 4.
Michael’s mother and father had different ideas about training. How did these two approaches help Michael fulfil his dream?
Answer:
Michael’s parents had different ideas about training. Mother encouraged the much needed dreaming. And father demanded the essential hard work. Mother provided inspiration. Father brought out perspiration (hard work). Thus Michael derived help from his parents.

Question 5.
What was so special about Michael’s achievement?
Answer:
Michael’s achievement was really so special. Michael was blind. And yet he became an Olympic champion.

Vocabulary

I. Synonyms :

Use a thesaurus/dictionary to find out synonyms of these words from the biographical account you have read.

Word	Synonyms
quench
glamour
passion
persistence
arrogant
pounding

Answer:

Word	Synonyms
quench	satisfy, fulfil, appease, fill, gratify
glamour	attraction, appeal, charm, grace, radiance
passion	eagerness, enthusiasm, emotion, fervour, fire, intensity, desire, zeal
persistence	perseverance, dedication, diligence, tenacity
arrogant	proud, conceited, vain, haughty, insolent
pounding	racing, running, beating, battering

Note: Synonyms are words with similar not the same – meaning. Synonyms should belong to the same part of speech.

II. Suffixes :

Suffixes are groups of letters that are attached to words at their ends. (Groups of letters attached to words at their beginnings are called prefixes.) Suffixes generally change the parts of speech of the words to which they are attached.

Look at the following words taken from the passage.

Now, pick out the suffix from each word and form new words of your own in the table given below. One is done for you.

Answer:

III. Collocations :

A collocation is an arrangement of words or other elements, especially those that commonly cooccur.
e.g.: heart-felt congratulation.
There are certain word combinations.

• Adjective + Noun : e.g. bright/harsh/intense/strong light
• Verb + Noun : e.g. cast/ emit/give/provide/shed light
• Noun + Verb : e.g. light gleams/glows/shines
• Noun +Noun : e.g. a light source
• Preposition+ Noun : e.g. by the light of the moon
• Noun+ Preposition : e.g. the light from the window
• Quantifier+ Noun : e.g. (of) a beam / ray of light

Use a dictionary and write which word in column 1 can collocate with those in the next five columns. Put a tick (✓) mark in the relevant column. Use the apt collocations in your own sentences.

Answer:

Sentences using collocations :

1. The mother has heartfelt sympathy on her children.
2. I received heartfelt congratulations from my friends when I obtained good marks.
3. You have our deep sympathy on the loss of your father.
4. Becoming a police officer was my childhood dream.
5. He expresses his opinion on his marriage with Latha.
6. Expressing strong opinion on you is very difficult to me.
7. She received a warm welcome from her friends in America.

Grammar

Look at the following sentences taken from the reading passage:

1. As he raced down the golden-lined wheat fields, he would always outrun the locomotives passing by.
2. When he heard the singing of some distant robins in flight, he knew it was his time to fly.
3. Since the other vaulter had fewer misses, Michael needed to clear this vault to win.

In the above sentences the clauses with as and when denote time whereas since denotes reason. As and when refer to time and the other parts state what happened at that time. In the third sentence, the clause beginning with since states the reason and the other states the consequence.

The words as, when, since are conjunctions, which connect two sentences (clauses). The clauses containing these words are called Adverbial clauses. These adverbial clauses cannot stand independently, so they are called Dependent clauses orSubordinate clauses. The clause that stands on its own is an Independent clause or Main clause.

The linkersas, when and since (Adverbs) are placed before the clauses to make them Subordinate clauses.

Now, read the biographical account once again and pick out the Adverbial clauses and Main clauses and write them in the table or in your notebook.

Answer:

Complete the following sentences with appropriate Adverbial clauses. Use the adverbs wherever they are given in brackets.

1. The teacher entered the class ___________ .
2. The dog ran into a speeding car ___________ .
3.1 was in deep sleep ___________ .
4. Someone knocked at the door ___________ .
5. The crowd cheered Michael Stone ___________ . (as soon as)
6. It is high time the cricket board thought of different alternatives ___________ . (as)
Answer:

1. The teacher entered the class when the bell had rung.
2. The dog ran into a speeding car as it did not notice the car.
3. I was in deep sleep when father came home and tried to wake me up.
4. Someone knocked at the door as I was doing my homework.
5. The crowd cheered Michael Stone as soon as he cleared the 17 feet 6½ inches height.
6. It is high time the cricket board thought of different alternatives as there have been a series of failures on the part of the Indian cricket team.

Writing

You have read the biography of Michael Stone. Think of some world-famous sporting personalities from India. Collect information about any one of them and write a biographical sketch.
You may include the following:
1. Date and place of birth
2. Information about the family
3. Achievements of the person : awards, prizes, honors, etc.
4. Important events in the life of the person – education, marriage, profession, etc.
5. Inspiration to others/message to the society
6. Contribution to his/her field and society
Answer:
Abhinav Bindre – The Ace Shooter

India’s thirst for a gold in Olympics individual events remained unquenched for over a hundred years. India had a taste of the individual gold only in 2008 Beijing Olympics when Abhinav Bindre achieved the land mark in 10M Air Rifle Shooting.

Abhinav Singh Bindre was born on 28 September 1982 in Dehradun in Uttarakhand (then Uttar Pradesh). Born into a very wealthy family of Sikhs, he had his early education in the world famous Doon School. Later, he moved to St. Stephen’s School from where he graduated in 2000. He did his B.B.A. in Colorado University of the U.S.A. He did a Diploma course in Mental Management too. He was pursuing his Doctorate in Business Management.

His skill as an expert shooter was identified quite early. His very supportive and affluent parents had got built exclusively for Bindre an indoor shooting range at home in Patiala! Services of a Mental coach Surabh Bhattacharjee, Lt. Col. Dhillon, and five time Olympic shooter Gabriele Buhlmans were made available to young Bhindre.

With his innate skills and excellent support from family, Bhindre started to shine from a very young age. At 15, he was the youngest participant in 1998 Commonwealth Games. His successes since then have been continuous and remarkable. He won a Bronze medal in the 2001 Munich World Cup. But in the same year, he won six Gold medals in various international events. 2002 Manchester Commonwealth Games crowned him with another Gold. 2004 Olympics saw him break the previous records, though he failed to win a medal. And in 2006, he became the first Indian to win the World Championship. 2006 Melbourne Commonwealth Games added one more Gold to his tally. Severe backache didn’t deter him from winning an Olympic Gold in 2008.

Various State Governments and Central Government honoured him amply and rewarded him with cash prizes. Arjuna Award in 2000 and Rajiv Gandhi Khel Ratna in 2001 are remarkable honours for him. Most memorable for him must be his flag bearing role in the opening ceremony of 2010 Commonwealth Games and taking the Athlete’s oath on behalf of 6,700 participants from 71 countries.

A Shot at History – My Obsessive journey to Gold is his version of his achievements. Abhinav Bindre’s life and achievements remain a source of inspiration to young aspirants.

Study Skills

You have finished reading the text “True Height”. Write the summary of the story in your own words. Use the following ideas to complete your summary effectively.

• Make a note of the important points in the text and the supporting details.
• Sequence of the events.
• Identify the words/ phrases which carry ideas.
• Use appropriate linkers.
• Focus on the words/phrases that express the essence of the text.
• Present the ideas briefly in your own words.
• Do not include examples in the summary.

Summary of The Reading Text

“True Height’ is a truly touching story of Michael Stone, written by David Naster. Michael was the son of Bert and Mildred Stones. Mildred always told Michael stories of flying high. That made Michael dream and dream high. Father Bert was a practical man. He always wanted Michael to work hard to win. He was also a coach and trainer to his son. With inspiration from mother, guidance from father, and his own determination, dedication, and discipline, Michael grew into an excellent pole-vault expert with the flexibility of an athlete and the strength of a body builder. Michael was also a good student and helped parents in their farm work. With his regular practice, devoted work he went on reaching unbeatable heights in pole-vault. By clearing 17 feet 6½ inches Michael created National and International Junior Olympics record. Congratulations overwhelmed Michael. Father was crying the greatest tears of pride. Michael Stone’s success was very significant. The reason … Michael Stone was blind.

Listening

Practise listening carefully. Then you will be able to speak.
Listen to the commentary on a cricket match and answer the questions.

Cricket Commentary

Hello, good morning, viewers. This is Sunil Gavaskar with Ravi Sastry. Welcome to Uppal Cricket Stadium, Hyderabad. It is a sunny morning. Electrifying atmosphere in this jam-packed Stadium. The crowd is expecting an exciting match-a war of nerves between the arch rivals, India and Pakistan. The local boy V.V.S. Laxman is the centre of attraction as it is his home ground.

Having won the toss, India elected to bat first. The inform openers Gambhir and Sehwag are at the crease.

The first ball of the day, an in-swinging yorker into the pads of Sehwag from Umar Gul. He is all hands up. There is a loud appeal for Ibw. Luckily, for Sehwag and for India it is a noball. There is absolute silence in the stadium.

Here comes Gul again! Sehwag is ready. Gul pitches the ball at the good length area. Sehwag goes backfoot, plays it gently to the square leg area. The batsmen cross for a quick single. Sehwag opens his account with that single. India is 2 for no loss of a wicket.

This brings Gambhir on to strike. The third ball of the over. This time a slightly misdirected ball on the leg stump and Gambhir comes forward and hits straight over the head of the bowler for a four.

Gul changes the guard. This time a very well-directed short-pitch delivery. Oh! What a delivery. But Gambhir connects the ball. What a sweet timing! In a flash the ball disappears into the stands. The umpire Steve Buckner declared it a six. The crowd jump on to their feet. With that Gambhir quickly moves on to 10 and India 12 without loss. There is a conversation between Gul and Shahid Afridi. It seems the captain has a piece of advice. And there is some field adjustment for Gambhir, Now let’s see how it works.

Umar Gul comes in and bowls to Gambhir. What a cracking delivery! It breaks the defense of Gambhir, Oh! The middle stump goes cart-wheeling. He’s out. Pakistan celebrates. The jubilant Gul runs to hug his captain. The crowd is dumbstruck. But the umpire Buckner signalled it a noball. Gambhir survives. A great sigh of relief on his face. How lucky this man is! The crowd comes alive again and cheers up. What an excitement in the very first over itself!

The final ball of the over. This time it is a slow ball from Gul. Gambhir elegantly pushes the ball to the onside and completes the run but there is a fumble by the fielder. The batsmen try for the second. Will they complete it? Yes. Now the score moves on to 15. Gambhir is on 12.

I. Tick (✓) only the correct ones from the statements given below.

1. Sachin and Gambhir opened the Indian innings. (✗)
2. Umar Gul opened the Pakistan’s attack. (✓)
3. Laxman is one among the Indian squad. (✓)
4. Gul bowled out Gambhir. (✗)
5. India is 15 without loss after the first over, (✓)

II. Answer the following questions.

1. Which two world teams of cricket do you like? Why?
Answer:
I like Indian team and Australian team, I love their art of playing, team spirit, sporting spirit and human approach.

2. Who are the openers?
Answer:
Sehwag and Gambhir are the openers.

3. Who are the commentators?
Answer:
Sunil Gavaskar and Ravi Sastri are the commentators.

III. Complete the Score Board given below.
India Vs Pakistan

Answer:

Oral Activity

Imagine, you were listening to the commentary, your father came there and reminded you of the ensuing examinations. He said, “Listening to cricket commentary is a waste of time.” Now develop a conversation between you and your father.
Answer:
Father : Hey Avinash ! What are you doing ? Examinations are just a week ahead, and you are listening to cricket commentary ! Isn’t it a sheer waste of time ?

Avinash : Sorry father! I just started to listen to commentary. I have been studying lessons for long. As I felt like relaxing for a few moments, I switched on the radio.

Father : But you can listen to music to ralax.

Avinash : True, father. But cricket commentary can also be useful in many ways. First, it gives us a chance to improve listening skills. Then it gives us quite interesting information. Analytical and presentation skills can also be improved. Sometimes we feel inspired to achieve great goals too I

Father : Sounds you are right. But I have my own doubts.

Avinash : Your doubts have base father. If we become an addict to commentary, we will waste all our time. As long as we are within our limits, listening to cricket commentary can be a productive and enjoyable activity.

Father : Seems we elders can pick up a point or two from you brilliant boys I

Avinash : Oh father, thanks a lot. But I will be in my limits.

Father : Good, son ! Continue to progress and prosper!

Avinash : I will try my level best father!

True Height Summary in English

“True Height’ is a truly touching story of Michael Stone, written by David Naster. Michael was the son of Bert and Mildred Stones. Mildred always told Michael stories of flying high. That made Michael dream and dream high. Father Bert was a practical man. He always wanted Michael to work hard to win. He was also a coach and trainer to his son. With inspiration from mother, guidance from father and his own determination, dedication and discipline, Michael grew into an excellent pole-vault expert with the flexibility of an athlete and the strength of a body builder. Michael was also a good student and helped parents in their farm work. With his regular practice, devoted work he went on reaching unbeatable heights in pole-vault. By clearing 17 feet 6^ inches Michael created National and International junior Olympics record. Congratulations overwhelmed Michael. Father was crying the greatest tears of pride. Michael Stone’s success was very significant. The reason … Michael Stone was blind.

True Height Glossary

height (n) : an extreme example of a particular quality
Note : Be careful about the spelling. It is h-e-i-g-h-t; not ‘high (adj) +1.’ And the letters ‘gh’ are silent.

athlete (n) : a person competing in sports

pursue (v-pre tense) : try to achieve

nerve-wracking (adj) : exciting; making one tense

ultimate (adj) : final, last, best

career(n) : job or work in a particular area

awe-inspiring (adj) : impressive; admiring

breathless (adj) : making one react emotionally

quenched (v-past tense) : satisfied

intensity (n) : a great extent; depth

astroturf (n) : artificial grass surface used in sports

confronted (v-past tense) : faced

pole vault (n) : a game in which a person jumps high with the help of a pole (a long stick)

gymnast (n) : a person participating in activities/sports that need strength and flexibility

fantasy (n) : a pleasant situation only in imagination

quest (n) : a long search

bird’s eye view (idiom) : a view from a high position

recurring (v-ing) : happening again and again

soaring (v-ing) : going up and up

coincide (v) : happen to be the same

hard-core (adj) : unchanging, firm in belief, strong

regimented (adj) : strictly disciplined

coach (n) : a person who guides, teaches techniques of a game

chores (n-plural) : works done regularly

persistence (n) : continuing to do something despite difficulties

obsession (n) : a state in which a person’s mind is completely filled with a thought of one particular thing/person.

arrogant (adj) : proud

inflated (adj) : filled with air

oblivious (adj) : not aware of; forgotten

surpassed (v – past tense) : exceeded; crossed

ritual (n) : a customary practice; formality

startled (v-past tense) : surprised; confused; threatened

bale of hay (n. phrase) : a bundle of dry grass

accurate (adj) : exact

envisioned (v-past) : saw in imagination

scared (adj) : afraid

a trickle (n) : drops in a series

pounding (v-ing) : beating quickly and noisily

The silence was deafening : The silence was unbearable
(Note the speciality of the expression – a paradox)

sprinting (v-ing) : running fast

eruption (n) : sudden, strong expression of powerful feelings by shouting

swarmed with (phrase) : surrounded by

accomplishment (n) : achievement

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles InText Questions

Question
Observe your surroundings carefully and write any three situations of your daily life where you can observe lines and angles. [Page No. 71]
Solution:
The edges of a blackboard; the edges of a ruler/scale and the edges of a table gives the idea of lines and their corners give the idea of angles.

Question
Draw the pictures in your notebook and collect some pictures. [Page No. 71]
Solution:

Think, Discuss and Write

What is the difference between inter-secting lines and concurrent lines ? [Page No. 74]
Solution:
If the number of lines meeting at a point are only two then they are said to be intersecting lines whereas if the lines are three or more than three meeting at a point then they are said to be concurrent lines.

Do These

Question 1.
Write the complementary, supplementary and conjugate angles for the following angles, a) 45° b) 75° c) 215° d) 30° e) 60° f) 90° g) 180° [Page No. 76]
Solution:

Question 2.
Which pairs of the following angles become complementary or supplementary. [Page No. 76]

Solution:
The angles formed by the figures (i) and (ii) are complementary.
The angles formed by the figures (ii) and (iii) are supplementary.

Try This

Question 1.
Find the pairs of adjacent and non- adjacent angles in the given figures. [Page No. 77]

Solution:
In figure (i) ∠1 and ∠2 are pair of adjacent angles.
In figure (ii), no adjacent angles.
In figure (iii) (∠1, ∠2), (∠2, ∠3) are pairs of adjacent angles.
In figure (iv) ∠1 and Z2 are adjacent angles.

ii) List the adjacent angles in the given figure. [Page No. 77]

Solution:
In the given figure
(∠1, ∠2), (∠3, ∠4), (∠4,∠5) and (∠3, ∠5) are the pairs of adjacent angles.

Think Discuss and Write

Question
Linear pair of angles are always supplementary. But supplementary angles need not form a linear pair. Why ? [Page No. 77]
Solution:
A pair of supplementary angles need not necessarily a linear pair because they may exists in separate figures.

Activity

Measure the angles in the following figure and complete the table.
[Page No. 78]

Solution:

Measure the four angles 1,2,3,4 in each of the above figure and complete the table:
[Page No. 79]

Solution:

Do This

Question 1.
Classify the given angles as pairs of complementary, linear pair, vertically opposite and adjacent angles. [Page No. 80]

Solution:
The pairs of angles a and b in the fig.(i) are linear pair of angles.
The pairs of angles a and b in the fig.(ii) are adjacent angles.
The pairs of angles a and b in the fig.(iii) are complementary angles*
The pairs of angles a and b in the fig.(iv) are vertically opposite angles.

Question 2.
Find the measure of angle ‘a’ in each figure. Give reasons in each case. [Page No. 81]
i)

a = 180° – 50° = 130°
(linear pair of angles)

ii)

a = 43°
( ∵ vertically opposite angles)

iii)

a = 360° – (209° + 96°)
= 360° – 305° = 55°
(∵ complete angle = 360°)

iv)

a = 90° – 63° = 27°
(pair of complementary angles )

Do These

Question 1.
Find the measure of each angle indi-cated in each figure where / and m are parallel lines intersected by a transversal n. [Page No. 87]

Solution:
x = 110° (alternate exterior angles)

Solution:
y = 84° (alternate interior angles)

Solution:
z = 180° – 100° = 80° .
(interior angles on the same side of transversal)

Solution:
s° = 53° (pair of corresponding angles)

Question 2.
Solve for x and give reasons. [Page No. 88]

11x + 2 = 75°
11x = 75 – 2 = 73
∴ x = \(\frac{73}{11}\)
(∴ pair of corresponding angles are equal).

Solution:
8x – 4 = 60°
8x = 60 + 4 = 64
∴ x = \(\frac{64}{8}\) = 8
(∴ alternate interior angles are equal)

Solution:
(14x- 1)°.= (12x + 17)°
14x – 12x = 17 + 1
2x = 18
x = \(\frac{18}{2}\) = 9
(∴ alternate exterior angles are equal)

Solution:
13x- 5 = 17x + 5
13x- 17x = 5 + 5
– 4x = 10
x = \(\frac{10}{-4}=\frac{-5}{2}\)
(∴ Pair of, corresponding angles are equal).

Activity

Question 1.
Take a scale and a ‘set sqaure’. Arrange the set sqare on the scale as shown in figure. Along the slant edge of set sqare draw a line with the pencil. Now slide your set square along its horizontal edge and again draw a line. We observe that the lines are parallel. Why are they parallel ? Think and discuss with your friends. [Page No. 88]
Solution:

A. Student Activity (For Reference .)
All slant lines are parallel making an angle of 60° with the horizontal line. In this figure horizontal line is transversal to the slant line and corresponding angles are equal.

Do This

Question
Draw a line \(\overline{\mathbf{A D}}\) and mark points B and C on it. At B and C, construct ∠ABQ and ∠BCS equal to each other as shown. Produce QB and SC on the other side of AD to form two lines PQ and RS.

Draw common perpendiculars EF and GH for the two lines PQ and RS. Measure the lengths of EF and GH. What do you observe ? What can you conclude from that ? Recall that if the perpendicular distance between two lines is the same, then they are parallel lines. [Page No. 89]


Solution:
As ∠ABQ =∠BCS and they lie on the same line AD we can say that BQ // CS. Now EF and GH are the perpendicular distances between two parallel lines PQ and R, we say EF = GH.

Try This

Question i)
Find the measure of the question marked angle in the given figure. [Page No. 90]

Solution:
? = 70°
[ ∵ from the figure, these two angles are exterior angles on the same side of the transversal]

ii) Find the angles which are equal to ∠P.
Solution:
∠P = ∠Q = ∠R = 110°
(corresponding angles)

Activity

Question
Draw and cut out a large triangle as shown in the figure.
Number the angles and tear them off.
Place the three angles adjacent to each other to form one angle as shown below. [Page No. 97]

1. Identify angle formed by the three adjacent angles ? What is its mea-sure ?
2. Write about the sum of the measures of the angles of a triangle. Now let us prove this statement
using the axioms; and theorems related to parallel lines.
Solution:
Student Activity.

Think, Discuss and Write

Question
If the sides of a triangle are produced in order, what will be the sum of exterion angles formed ? [Page No. 99]
Solution:
Let ΔABC and the sides of the triangle is formed by exterior angles.

∠3 = ∠B + ∠C
∠1 + ∠2 + Z∠3 = 2[∠A + ∠B + ∠C]
= 2 x 180° = 360°
∴ Sum of the exterior angles are 360°.

AP Board 9th Class Maths Solutions