Class 7

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 15 Symmetry Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 15th Lesson Symmetry Exercise 2

Question 1.
In the figures given below fmd the axes of symmetry such that on folding along the axis the two dots fall on each other.


Solution:

Question 2.
Given the line of symmetry, find the other dot.

Solution:

Question 3.
In the following incomplete figures, the mirror line (i.e. the line of symmetry) is given as a dotted line. Complete each figure, pcrlorming reflection on the dotted (mirror) line. (You might perhaps place a mirror along the dotted line and look into the mirror for the image). Are you able to recall the name of the figure you complete’?

Solution:

Question 4.
State whether the following statements are true or false.
(i) Every closed figure has an axis of symmetry. ( )
(ii) A figure with at least one axis of symmetry is called a symmetric figure. ( )
(in) A regular polygon of 10 sides will have 12 axes of symmetry. ( )
Solution:
(i) False
(ii) True
(iii) False

Question 5.
Draw a square and construct all its axes of symmetry. Measure the angles between each pair of successive axes of symmetry. What do you notice? Does the same rule apply for other regular polygons’?

Solution:
Angle between successive axes of
Symmetry = 45° = \(\frac{360^{\circ}}{2 \mathrm{n}}\)
This is true for all regular polygons.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 15 Symmetry Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 15th Lesson Symmetry Exercise 1

Question 1.
Given below are some fiugres. Which of them are symmetric? Draw the axes of symmetry
for the symmetric figures.

Solution:

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 14th Lesson Understanding 3D and 2D Shapes Exercise 4

Question 1.
A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. Attempt to give a rough sketch of the shadow. (You may try to experiment first and then answer these questions).

Solution:
A ball -Circle
A cylindrical pipe – Rectangle
A book – Rectangle

Question 2.
Here are the shadows of some 3D objects, when seen under the lamp of an overhead projector. Identify the solid(s) that match each shadow. (There may be many answers for these!)

Solution:
i) A Circle – Spherical/circular objects
ii) A Square – Cube/square sheets
iii) A Triangle – Triangular/right prism with triangular base
iv) A Rectangle – Cuboid/rectangular shapes.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 14th Lesson Understanding 3D and 2D Shapes Exercise 3

Question 1.
Use an isometric dot paper and make an isometric sketch for each one of the given shapes.

Solution:

Question 2.
The dimensions of a cuboid are 5 cm, 3 cm and 2 cm. Draw three different isometric sketches of this cuboid.
Solution:

Question 3.
Three cubes each with 2 cm edge are placed side by side to form a cuboid. Draw an oblique or isometric sketch of this cuboid.
Solution:

Question 4.
Make an oblique sketch for each of the given isometric shapes.

Solution:

Question 5.
Give (i) an oblique sketch and (ii) an isometric sketch for each of the following:
(a) A cuboid of dimensions 5 cm, 3 cm and 2 cm. (Is your sketch unique’?)
(b) A cube with an edge 4 cm long.
Solution:

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 14th Lesson Understanding 3D and 2D Shapes Exercise 2

Question 1.
Some nets are given below. Trace them and paste them on a thick paper. Try to make 3-D
shapes by suitably folding them and gluing together. Match the net with it’s 3-D shape.
Nets
3D shapes
(i)

Solution:

Question 2.
Three nets for each shape are given here. Match the net with its 3-D shape.
Solution:
Solution:


Solution:

Question 3.
A dice is a cube with dots on each face. The opposite faces of a dice always have a total of seven dots on them.
Here are two nets to make dice. Insert the suitable number of dots in blanks.

Solution:

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 14th Lesson Understanding 3D and 2D Shapes Exercise 1

Question 1.
Given below are the pictures of some objects. Categorise and fill write their names according to their shape and fill the table with name of it.

Solution:

Question 2.
Write names of at least 2 objects from day-to-day life, which are in the shape of the basic 3D shapes given below:
(i) Cone – ……………
(ii) Cube – ……………
(iii) Cuboid – ……………
(iv) Sphere – ……………
(v) Cylinder – ……………
Solution:
i) Cone : Ice-cream cone, birthday caps
ii) Cube : Die, chalk piece box
iii) Cuboid : Eraser, brick
iv) Sphere : Marbles, balls
v) Cylinder: Pencil, candle, pipe

Question 3.
Identify and state the number of faces, edges and vertices of the figures given below:

Solution:

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 6 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 6

Question 1.
A path 2.5 m wide is running around a square field whose side is 45 m. Determine the area of the path.
Solution:

Area of the path= (Area of outer figure) – (Area of inner figure)
= (50 × 50) – (45 × 45)
= 2500 – 2025 = 475 m²

Question 2.
The central hail ofa school is 18m long and 12.5 m wide. A carpet is to be laid on the floor leaving a strip 50 cm wide near the walls, uncovered. Find the area of the carpet and also the uncovered portion?
Solution:

Given
Length of the hail = 18m
Breadth of the hail 12.5m
Width of the strip = 50cm = \(\frac { 1 }{ 2 }\)m
Length of inner rectangle = 18 – (\(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 2 }\)) 17m
Breadth of inner rectangle = 12.5 – (\(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 2 }\)) = 11.5 m
Area of the strip = (Area of the outer fIgure) – (Area of the inner figure)
= 18 × 12.5 – 17 ×11.5
= 225 – 195.5 = 29.5 m²
∴ Area of the carpet = 195.5 m²

Question 3.
The length of the side of a grassy square plot is 80 m. Two walking paths each 4m wide
are constructed parallel to the sides of the plot such that they cut each other at the centre
of the plot. Determine the area of the paths.
Solution:

Given : A square grassary plot of length 80m
Two paths of width 4m each.
From the question
KL = 4m and KN = 80m
EH = 4m and EF 80m
PQ = 4m and PS 4m
Area of two paths = 🖾 KLMN + 🖾 EFGH – 🖾 FQRS
= KN x KL + EF × EH – PQ × PS
= 80 × 4 + 80 × 4 – 4 × 4
= 320 + 320 – 16
= 624sq.m.

Question 4.
A verandah 2 m wide is constructed all around a room of dimensions 8 m X 5 m. Find the area of the verandah
Solution:

Length of outer rectangle = 8 + 2 + 2 =12m
Breadth of outer rectangle = 5 + 2 + 2 = 9m
Area of outer rectangle = l x b = 12 × 9 = 108m²
Length of inner rectangle = 8m
Breadth of inner rectangle = 5m
Area of inner rectangle = l x b = 8 × 5 = 40m²
:. Area of the verandah = (Area of outer figure) – (Area of inner figure)
= 108 – 40 = 68m²

Question 5.
The length of a rectangular park is 700 m and its breadth is 300 m. Two crossroads, each of width 10 m, cut the centre of a rectangular park and are parallel to its sides. Find the area of the roads. Also, find the area of the park excluding the area of the crossroads.
Solution:

Given
Length of the rectangular park = 700 m
Breadth of the rectangular park = 300 ni
From the question,
PS 10m. PQ = 700 ni
Area of 🖾 PQRS = 10 × 700 = 7000m²
AB = 10m, AD = 300m
Area of 🖾 ABCD = 300 x 10 3000m²
KL = 10m, KN = 10m
🖾 KLMN = 10 × 10 = 100m²
Area of the two paths = 🖾 IQRS + 🖾 AF3CD – 🖾 KLMN
= 7000 + 3000 – 100
= 9900m²
The remaining area of the park = Area of the whole park – Area of the two paths
= 700 × 300 – 9900
= 210000 – 9900
= 200100 q.m.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 7 Data Handling Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Data Handling Exercise 4

Question 1.
Draw a bar graph for the following data.
Population of India in successive census years-

Solution:

Question 2.
Draw a pie chart for the following data.

Solution:

Question 3.
Draw a double bar graph for the following data.
Birth and Death rates of different states in 1999.

Solution:

Question 4.
Draw a pie chart for the following data.
Time spent by a child during a day-

Solution:

Question 5.
The adjoining pic chart gives the expenditure on various items during a month for a family.
(The numbers written around the pie chart tell us the angles made by each sector at the centre.)

Answer the following –
(i) On which item is the expenditure minimum?
(ii) On which item is the expenditure maximum?
(iii) If the monthly income of the family is ₹ 9000, what is the expenditure on rent?
(iv) If the expenditure on food is ₹ 3000, what is the expenditure on education of children?
Solution:
i) Education
ii) Food
iii) Total income is represented by 360° = Rs. 9,000
food represented by 120 = \(\frac{120}{360}\) × 9000 = Rs. 3000
iv) The expenditure on food is represented by 120 = Rs. 3000
Then expenditure on education is represented by \(\frac{60}{120}\) × 3000 = Rs. 1500

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 7 Data Handling Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Data Handling Exercise 3

Question 1.
Say true or false and why?
(i) The difference between the largest and smallest observations in a data set is called the
mean.
(ii) In a bar graph, the bar which has greater length indicates mode.
(iii) Value of every observation in the data set is taken into account when median is calculated.
(iv) The median of a set of numbers is always one of the numbers
Solution:
i) False
ii) True
iii) False.
We take only the mid value when the observations are arranged either in ascending / descending order.
iv) False
It may or may not be in the data set.

Question 2.
The monthly income (in rupees) of 7 households in a village are 1200, 1500, 1400, 1000, 1000, 1600, 10000. (i) Find the median income of the house holds. (ii) If one more household with monthly income of 1500 is added, what will the median income be?
Solution:
Given that
The household incomes in Rs. are 1200, 1500. 1400, 1000, 1000, 1600, 10,000
Arranging these observations in ascending order 1000, 1000, 1200, 1400, 1500, 1600, 10,000
Medían is the mid value = Rs. 1400
If one more household with income Rs. 1500 ¡s added them the data becomes
1000, 1000, 1200. 1400, 1500, 1500, 1600, 10,000
Now the median = average of 1400 and 1500
= \(\frac{1400+1500}{2}=\frac{2900}{2}\) = Rs.1450

Question 3.
Observations of a data are 16, 72,0, 55, 65, 55, 10, and 41. Chaitanya calculated the mode and median without taking the zero into consideration. Did Chaitanya do the right thing?
Solution:
Mode is correct but median is wrong.

Question 4.
How many distinct sets of three positive integers have a mean of 6, a median of 7, and no mode?
Solution:
It is not possible

Question 5.
Four integers arc added to a group of integers 3,4,5,5 and 8 and the mean, median, and mode of the data increases by 1 each. What is the greatest integer in the new group of integers?
Solution:
Given set of integers = 3, 4, 5, 5 and 8
The mean of the given data = \(\frac{\text { Sum of the integers }}{\text { Number of integers }}=\frac{3+4+5+5+8}{5}=\frac{25}{5}\) = 5

The median of the given data 3. 4. 5, 5. 8 (already in ascending order) = 5 (the mid
Mode of the given data 3. 4. 5. 58 iS 5.

After adding 4 integers the mean, mode and median increased by 1
New
Mean = 5 + 1 = 6
Median = 5 + 1 = 6
Mode = 5 + 1 = 6

Now sum of the (5 + 4 = 9) numbers = Number of integers × average
Sum = 9 × 6 = 54
But sum of the given set of 5 numbers = 25
∴ Sum of the newly added 4 numbers = 54 – 25 = 29
Mode = 6 means 6 should appear for 3 times.
But sum of four numbers = 29
29 = 6 + 6 + 6 + fourth number
29 = 18 + fourth number
∴ fourth number = 29 – 18 = 11

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 5 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 5

Question 1
Find the circumference of a circle whose radius is
(i) 35cm (ii) 4.2cm (iii) 15.4 cm
Solution:
i) Given: r= 35 cm
Circumference of the circle = 2πr = 2 × \(\frac{22}{7}\) × 35 = 220 cms.

ii) Given : r = 4.2 cm
Circumference of the circle = 2πr = 2 × \(\frac{22}{7}\) × 4.2 26.4 cms

iii) Given: r = 15.4cm
Circumference of the circle = 2πr = 2 × \(\frac{22}{7}\) × 15.4 = 96.8 cms

Question 2.
Find the circumference of circle whose diameter is
(i) 17.5 cm (ii) 5.6 cm (iii) 4.9 cm
Note : take π = \(\frac{22}{7}\) in the above two questions.
Solution:
i) Given : d = 17.5 cm
Circumference of the circle = πd = \(\frac{22}{7}\) × 17.5 = 55 cm
ii) Given d = 5.6 cm
Circumference of the circle = πd = \(\frac{22}{7}\) × 5.6 = 17.6 cm
iii) Given d = 4.9 cm
Circumstances of the circle = πd = \(\frac{22}{7}\) × 4.9 = 15.4 cm

Question 3.
(i) Taking π = 3.14, find the circumference of a circle whose radius is
(a) 8 cm (b) 15 cm (c) 20cm
(ii) Calculate the radius of a circle whose circumference is 44cm?
Solution:
i) a) Given r = 8 cm π = 3.14
Circumference of circle = 2πr
= 2 × 3.14 × 8 = 50.24cm

b) Given r = 15cm
Circumference of circle = 2πr
= 2 × 3.14 × 15 = 942cm

c) Given r 20cm
Circumference of circle = 2πr
= 2 × 3.14 × 20 = 125.6cm

ii) Given,
Circumference = 44cm
Circumference of a circle = 2πr
2 × \(\frac{22}{7}\) x r =
2 × 22 × r = 44 x 7
r = \(\frac{44 \times 7}{2 \times 22}\) = 7cm
∴ radius of the circle = 7cm.

Question 4.
If the circumference of a circle is 264 cm, find its radius. Take π = \(\frac{22}{7}\)
Solution:
Given : Circumference = 264 cm
Circumference of a circle = 2πr = 2 × \(\frac{22}{7}\) x r = 264
2 × 22 × r = 264 × 7
r = \(\frac{264 \times 7}{2 \times 22}\) = 42
∴ radius of the circle = 42 cm.

Question 5.
If the circumference of a circle is 33 cm, fmd its diameter.m
Solution:
Given circumference = 33cm
Circumference of a circle = πd
\(\frac{22}{7}\) × d = 33
22d = 33 × 7
d = \(\frac{33 \times 7}{22}=\frac{21}{2}\) = 10.5cm
∴ diameter of the circle = 10.5 cm

Question 6.
How many times will a wheel of radius 35cm be rotated to travel 660 m?
(Take π = \(\frac{22}{7}\)).
Solution:
Given radius = 35cm.
Circumference of the circle = 2πr
= 2 × \(\frac{22}{7}\) × 35
= 220cm.
∴ The number of times the wheel will he rotated = \(\frac{660}{220}\) = 3 times.

Question 7.
The ratio of the diameters of two circles is 3 :4. Find the ratio of their circumferences.
Solution:
Given: Ratio of diameter = d₁ : d₂ = 3 : 4
Ratio of their circumferences C₁ : C₂ = πd₁ : πd₁
= \(\frac{22}{7}\) × 3: \(\frac{22}{7}\) : 4 = 3 : 4

Question 8.
A road roller makes 200 rotations in covering 2200 m.
Find the radius of the roller.
Solution:
Given : Road roller makes 200 rotations covering a distance = 2200m
(i.e.,) 200 × circumference = 2200
200 × 2 × π × r = 2200
200 × 2 \(\frac{22}{7}\) × r = 2200
r = 2200 × \(\frac{7}{22} \times \frac{1}{200} \times \frac{1}{2}=\frac{7}{4}\) = 1.75m

Question 9.
The minute hand of a circular clock is 15 cm.
How far does the tip of the minute hand move in 1 hour?
(Take π = 3.14)
Solution:
Given : The length of minute hand = 15cm
In one hour it completes a rotation.
i.e. distance covered by its tip = circumference of the circle with a radius 15cm
22 660
= 2 × \(\frac{22}{7}\) × 15 = \(\frac{660}{7}\) cm = 94.28cm

Question 10.
A wire is bent in the form of a circle with radius 25 cm. It is straightened and made ¡rito a square. What is the length of the side of the square?
Solution:
Given: Radius of the circle 25 cm .
Circumference of the circle = Perimeter of the square
2πr = 4 × side
4 × side=2 × \(\frac{22}{7}\) × 25
Side = 2 × \(\frac{22}{7}\) × 25 × \(\frac{1}{4}\) = \(\frac{275}{7}\) = 39.28
∴ Side of square = 39.28 cm

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 7 Data Handling Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Data Handling Exercise 1

Question 1.
Maximum day time temperatures of Hyderabad in a week (from 26th February to 4th March, 2011) are recorded as 26°C, 27°C, 30°C, 30°C, 32°C, 33°C and 32°C.
(i) What is the maximum temperature of the week?
(ii) What is the average temperatures of the week?
Solution:
i) Maximum temperature = 33C
ii) Average temperature = \(\frac{\text { Sum of the temperatures }}{\text { No. of observations }}\)
= \(\frac{26+27+30+30+32+33+32}{7}=\frac{210}{7}\)
∴ The average temperature of the week 30°C.

Question 2.
Rice consumed in a school under the mid-day meal program for 5 consecutive days is 15.750 kg, 14.850 kg, 16.500 kg, 14.700 kg, and 17.700 kg. Find the average rice consumption for the 5 days.

Solution:
Rice consumed for 5 days in kg = 15.750, 14.850, 16.500 14.700, 17.700

∴ The average rice consumption for the 5 days = 15.900 kg.

Question 3.
In a village three different crops are cultivated in four successive years. The profit (in rupees) on the crops, per acre is shown in the table below-

(i) Calculate the mean profit for each crop over the 4 years.
(ii) Based on your answers, which crop should be cultivated in the next year?
Solution:

(ii) As the mean profit on Groundnuts is more than the other two crops, Groundnuts may be cultivated for the next year.

Question 4.
The number of passengers who travelled in APSRTC bus from Adilabad to Nirmal in 4 trips in a day are 39, 30, 45 and 54. What is the occupancy ratio (average number of passengers travelling per trip) of the bus for the day?

Solution:
Passengers travelled in 4 days 39. 30, 45 and 54
Average / Occupancy ratio = \(\frac{\text { Sum }}{\text { Number }}=\frac{\text { Total passengers travelled }}{\text { Number of days }}\)
= \(\frac{39+30+45+54}{4}=\frac{168}{4}=42\)
∴ The occupancy ratio of the bus for the day = 42

Question 5.
The following table shows the marks scored by Anju, Neelesh and Lckhya in four unit tests of English.

(i) Find the average marks obtained by Lekhya.
(ii) Find the average marks secured by Anju. Will you divide the total marks by 3 or 4?Why?
(iii) Neelesh has given all four tests. Find the average marks secured by him. Will you divide the total marks by 3 or 4? Why?
(iv) Who performed best in the English?
Solution:
Average = \(\frac{\text {Sum of the observations }}{\text { Number of observations }}\)
i) Average marks obtained by Lekhya = \(\frac{20+24+24+24}{4}=\frac{92}{4}\) = 23
ii) Anju has given only three tests. So to find the average marks we divide by 3.
(i.e.) average = \(\frac{19+23+21}{3}=\frac{63}{3}\) = 21
iii) Neelesh has given 4 tests. So to find the average marks we divide by 4.
(i.e.) average = \(\frac{0+20+22+24}{4}=\frac{66}{4}\) = 16.5
iv) As the average marks of Lekhya is greater than the other two, we conclude that Lekhya
performed best in English.

Question 6.
Three friends went to a hotel and had breakfast to their taste, paying 16, 17 and 21 respectively
(i) Find their mean expenditure.
(ii) If they have spent 3 times the amount that they have already spent, what would their mean expenditure be?
(iii) If the hotel manager offers 50% discount, what would their mean expenditure be?
(iv) Do you notice any relationship between the change in expenditure and the change in mean expenditure.
Solution:
i) Mean expenditure = \(\frac{\text { Total expenditure }}{\text { No. of persons }}\)
= \(\frac{16+17+21}{3}=\frac{54}{3}\) = Rs. 18

ii) Amount spent = 3 times I.e., 3 × 16; 3 × 17; 3 × 21
= Rs. 48; Rs. 51; Rs. 63
Now the average = \(\frac{48+51+63}{3}=\frac{162}{3}\) = Rs. 54 = 3 x original average

iii) After 50% discount the amount spent is Half of the actual amount = \(\frac{16}{2}, \frac{17}{2}, \frac{21}{2}\)
= Rs.8; Rs. 8.50; Rs. 10.50
Now the average = \(\frac{8+8.50+10.50}{3}=\frac{27.00}{3}\) = Rs .9 = \(\frac{\text { Original average }}{2}\)

iv) The change In the observations is also carried out in the mean.

Question 7.
Find the mean of the first ten natural numbers.
Solution:
First ten natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10
∴ Average = \(\frac{\text { Sum of the numbers }}{\text { Number }}\)
= \(\frac{1+2+3+4+5+6+7+8+9+10}{10}=\frac{55}{10}\) = 5.5
∴ The mean of the first ten natural numbers = 5.5.

Question 8.
Find the mean of the first five prime numbers.
Solution:
First five prime numbers are 2, 3, 5, 7 and 11
Average of first five prime numbers = \(\frac{\text { Sum of the numbers }}{\text { Number of primes }}\)
= \(\frac{2+3+5+7+11}{5}=\frac{28}{5}\) = 5.6

Question 9.
In a set of four integers, the average of the two smallest integers is 102, the average of the three smallest integers is 103, the average of all four is 104. Which is the greatest of these integers?
Solution:
Given that
The average of four lntegers = 104 = \(\frac{\text { Sum of the four integers }}{\text { Number of integers }}\)
∴ The sum of the four integers = average × number
= 104 × 4
= 413
Also the average of the three smallest integers = 103 = \(\frac{\text { Sum of the three smallest integers }}{\text { Number of integers }}\)
∴ The sum of the smallest three integers = average × number
= 103 × 3
= 309
∴ The greatest integer / fourth number (Sum of four Integers) – (Sum of three integers)
=416 – 309 = 107

Question 10.
Write at least two questions to find the mean, giving suitable data.
Solution:
Q – 1: The daily income of a shop-keeper during 6 days are Rs. 350, Rs. 325, Rs, 400, Rs. 450, Rs. 600, Rs. 120. Find his average daily income.
Q – 2: The number of eggs sold by a poultry during 5 days are 480, 512, 680, 720 and 1026. Find the average daily sales.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 4

Question 1.
Find the area of the following rhombuses.

Solution:
Area = \(\frac { 1 }{ 2 }\)d₁d₂
d₁ = 5 + 5 = 10cm
d₂ = 2 + 2 = 4cm
A = \(\frac { 1 }{ 2 }\) × 10 × 4 = 20 cm²

d₁ = 3 + 3 = 6 cm
d₂ = 4 + 4 = 8cm
Area = \(\frac { 1 }{ 2 }\)d₁d₂
= \(\frac { 1 }{ 2 }\) × 6 × 8
= 24 cm²

Question 2.
Find the missing values.

Diagonal – 1 (d1)	Diagonal – 2 (d2)	Area of the rhombus
12cm	16cm
27mm		2025mm2
14m	57.6m

Solution:

Diagonal – 1 (d1)	Diagonal – 2 (d2)	Area of the rhombus
12cm	16cm	\(\frac{1}{2}\) × 12 × 16 = 96cm2
27mm	\(\frac{2025 \times 2}{27}\) = 150mm	2025mm2
14m	57.6m	\(\frac{1}{2}\) × 24 × 57.6 = 691.2 m2

Question 3.
If length of diagonal of a rhombus whose area 216 sq. cm. is 24 cm. when find the length of second diagonal.
Solution:
Given: Length of one diagonal d₁ = 24cm, d₂ =?
Area = \(\frac { 1 }{ 2 }\)d₁d₂ = 216
\(\frac { 1 }{ 2 }\) × d₁ × d₂ = 216
\(\frac { 1 }{ 2 }\) × 24 × d₂ = 216
d₂ = \(\frac{216}{12}\) = 18

Question 4.
The floor of a building consists of3 000 tiles which are rhombus shaped. The diagonals of each of the tiles are 45 cm and 30 cm. Find the total cost of polishing the floor, if cost per m² is Rs.2.50.
Solution:
Diagonals of each (shape / rhombus) tiles
d₁ = 45cm, d₂ = 30cm
Total tiles = 3000
Total area = 3000 × Area of each tile = 3000 × \(\frac { 1 }{ 2 }\) × d₁ × d₂
= 3000 x \(\frac { 1 }{ 2 }\) × 45 × 30 = 2025000 cm²
= \(\frac{2025000}{100 \times 100}\) m² = 202.5m²
∴ Cost of polishing the floor at the rate of Rs. 2.50 per a square metre 202.5 × 2.50 = Rs. 506.25