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These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.4

Question 1.
A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also, find the area of the garden in hectare.
Answer:
Length of the garden (1) = 90 m
Breadth of the garden (b) = 75 m
Area of the garden = 1 x b sq units = 90 m x 75 m = 6750 m²

Length of the outer rectangle (L)
= 90 + 5 + 5 m
= 100 m
Breadth of the outer rectangle (B)
= 75 + 5 + 5 m
= 85 m
Area of the outer rectangle
= L x B = 100 x 85 m²
= 8500 m²
Area of the pathway = Area of the outer rectangle- Area of the inner rectangle
= (8500 – 6750)m²
= 1750m²
(i) Area of the garden = 6750 m²
= \(\frac{6750}{10000}\) ha
= 0.675 ha (1 m² = \(\frac{1}{10000}\) ha)
(ii) Area of the path = 1750 m²

Question 2.
A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
Answer:
Length of the park (1) = 125 m
breadth of the park (b) = 65 m
Area of the park = l x b
= 125 x 65 m²
= 8125m²

Length of the outer region (L)
= (125 + 3 + 3) m
= 131 m
Breadth of the outer region (B)
= (65 + 3+ 3) = 71 m
Area of the outer region
= L x B m²
= (131 x 71) m²
= 9301 m²
Area of the path = Area of the outer region – Area of the park
= 9301 m² – 8125 m²
= 1176 m²

Question 3.
A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Answer:
Length of the cardboard (l) = 8 cm
Width of the cardboard (b) = 5 cm
Area of the card board = 1 x b sq.
= 8 x 5 = 40 cm²

Width of the margin = 1.5 cm
Length of the inner rectangle
= 8 – (1.5 + 1.5) cm
= 5 cm
Breadth of the inner rectangle
= 5 – (1.5 + 1.5)cm
= 2 cm
Area of the inner rectangle
= 5 x 2 = 10 cm²
Area of the margin = Area of the cardboard – Area of the inner rectangle
= 40 cm² – 10 cm²
= 30 cm²

Question 4.
A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:
(i) the area of the verandah.
(ii) the cost of cementing the floor of the verandah at the rate of ₹ 200 per m² .
Answer:
Length of the room= 5.5 m
Breadth of the room = 4 m
Area of the room = 5.5 m x 4 m = 22 m²
Width of the verandah = 2.25 m
Length of the verandah
= 5.5 + (2.25 +2.25) m
= 10 m

Breadth of the Verandah
= 4 + (2.25 + 2.25) m
= 8.5 m
Area of the outer rectangle
= 10 m x 8.5 m ‘
= 85 m²
Area of the verandah = Area of the outer rectangle – Area of the inner rectangle
= 85 m² – 22 m²
= 63 m²
Cost of cementing the verandah
= ₹ 200 x 63
= ₹ 12600

Question 5.
A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:
(i) the area of the path
(ii) the cost of planting grass in the remaining portion of the garden at the rate of? 40 per m².
Answer:
(i) Length of the outer square = 30 m
Area of the outer square = side x side
= 30 x 30 = 900 m²

Let width of the path = 1 m
Side of the inner square
= [30 – (1 + 1)3 m
= 30m-2m = 28m
Area of the inner square
= (28 x 28) m²
= 784 m²
Area of the path = Area of the outer square – Area of the inner square
= (900 – 784) m²
Area of the path = 116 m²

(ii) Rate of planting grass
= ₹ 40 per m²
Cost of planting grass = ₹ 40 x 784
= ₹ 31,360

Question 6.
Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.
Answer:
Length of the rectangular park (l) = 700 m
Breadth of the rectangular Park (b) = 300 m

Area of the park = l x b sqm
= 700 x 300 m²
= 210000 m²
Area of the road HEFG (length wise)
= 700 m x 10 m
= 7000 m²
Area of the road PQRS (breadth wise)
= 300 x 10 = 3000 m²
Area of KLMN = 10 x 10 = 100 m²
Area of the roads = Area of the road HEFG + Area of the Road PQRS – Area of KLMN (which is repeated two times.)
= (7000 +3000-100) m²
= 9900 m²
Area of the park excluding cross roads = Area of the park – Area of the cross roads
= 21,0000 m² – 9900 m² = 200100 m²
= \(\frac{200100}{10000}\) ha
∴ Area of the park = 20.01ha

Question 7.
Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find
(i) the area covered by the roads.
(ii) the cost of constructing the roads at the rate of? 110 per m2.
Answer:
Length of the rectangular field = 90 m
Breadth of the rectangular field = 60 m
Width of each road = 3m.

Area of the road ABCD = 90 x 3 m²
= 270 m²
Area of the road EFGH = 60 x 3
= 180 m²
Area of PQRS = 3 x 3 = 9 m²
(i) Area covered by the roads = Area of the road ABCD + Area of the road EFGH – Area of PQRS (which is repeated two times)
= (270 + 180 – 9) m²
= 450 m² – 9 m²
= 441 m²
(ii) Rate of construction of roads = ₹ 110/m²
Cost of construction of roads
= ₹ 110 x 441
= ₹ 48510

Question 8.
Pragya wrapped a cord around a circular pipe of radius 4 cm (figure given below) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (π = 3.14)
Answer:
Radius of the circular pipe (r) = 4 cm
Circumference of the pipe = 2πr.
= 2 x 3.14 x 4 cm = 25.12 cm

Side of the square box = 4 cm
Perimeter of the square box
= 4 x 4 cm
= 16 cm
Since, 25.12 cm > 16 cm
Difference in length
= 25.12 cm – 16 cm
= 9.12 cm.
Yes, she has 9.12 cm of length cord left.

Question 9.
The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find: (Take π = 3.14)
(i) the area of the whole land
(ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower bed
(iv) the circumference of the flower bed.
In the following figure, find the area of the shaded portions:

Answer:
Length of the land = 10 m
Breadth of the land = 5 m
(i) Area of the whole land
= (10 x 5) m²
= 50 m²
(ii) Radius of the flower bed (r)
= 2 m.
Area of the flower bed = πr² sq.m
= 3.14 x 2 x 2 m² = 12.56 m²

(iii) Area of the lawn excluding the area of the flower bed = 50 m² – 12.56 m²
= 37.44 m²

(iv) Circumference of the flower bed
= 2πr.
= 2 x 3.14 x 2 m
= 12.56 m

Question 10.
In the following figures, find the area of the shaded portions.

Answer:
(i) Area of the whole rectangle ABCD
= 18 x 10 cm² = 180 cm²
Area of the right ΔAEF
= \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 6 x 10 cm²
= 30 cm²
Area of the right ΔCBE
= \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 8 x 10 cm²
= 40 cm²
Area of the shaded portion = Area of ABCD – (Area of ΔAEF + Area of ΔCBE)
= 180 – (30 + 40)cm² = 180 cm² – 70 cm² = 110 cm²
(ii) Side of the square PQRS = 20 cm
Area of the square PQRS
= Side x Side
= 20 x 20 cm²
= 400 cm²
Area of the right ΔQPT
= \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 20 x 10 cm²
= 100 cm²
Area of the rightΔTSU
= \(\frac { 1 }{ 2 }\) x 10 x 10 cm²
= 50 cm²
Area of the right AQRU
= \(\frac { 1 }{ 2 }\) x 10 x 20 cm²
= 100 cm²
Area of the shaded portion = Area of the square PQRS – (Area of ΔQPT + Area of ΔTSU + Area of ΔQRU)
= 400 cm² – (100 + 50 + 100) cm²
= (400 – 250) cm²
= 150 cm²

Question 11.
Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC
Answer:
Area of ΔABC

= \(\frac { 1 }{ 2 }\) x AC x BM
= \(\frac { 1 }{ 2 }\) x 22 x 3 cm²
= 33 cm²
Area of AACD = \(\frac { 1 }{ 2 }\) x AC x ND
= \(\frac { 1 }{ 2 }\) x 22 x 3 cm²
= 33 cm²
∴ Area of the quadrilateral ABCD = Area of ΔABC + Area of ΔACD
= 33 cm² + 33 cm²
= 66 cm²

These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.2

Question 1.
Find the square of the following numbers.
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46
Solution:
(i) 32²
= (30 + 2)²
= 30² + 2 (30) (2) + 2² [∵ (a + b)² = a² + 2ab + b² ]
= 900 + 120 + 4
= 1024

(ii) 35²
= (30 + 5)²
= 30² + 2 (30) (5) + 5²
= 900 + 300 + 25
= 1225

(iii) 86²
= (80 + 6)²
= 80² + 2 (80) (6) + 6²
= 6400 + 960 + 36
= 7396

(iv) 93²
= (90 + 3)²
= 90² + 2 (90) (3) + 3²
= 8100 + 540 + 9
= 8649

(v) 712
= (70 + 1)²
= 70² + 2(70) (1) + 1²
= 4900 + 140 + 1
= 5041

(vi) 46²
= (40 + 6)²
= 40² + 2 (40) (6) + 6²
= 1600 + 480 + 36
= 2116

Question 2.
Write a Pythagorean triplet whose one member is
(i) 6
(ii) 14
(iii) 16
(iv) 18
Solution:
(i) Let 2n = 6
n = \(\frac{6}{2}\) = 3

n² – 1
= 3² – 1
= 8

n² + 1
= 3² + 1
= 10
∴ The required Pythagorean triplet is 6, 8 and 10.

(ii) Let 2n = 14
n = \(\frac{14}{2}\) = 7

n² – 1
= 7² – 1
= 49 – 1
= 48

n² + 1
= 7² + 1
= 49 + 1
= 50
∴ Thus, the required Pythagorean triplet is 14, 48 and 50.

(iii) Let 2n = 16
n = \(\frac{16}{2}\) = 8

n² – 1
= 8² – 1
= 64 – 1
= 63

n² + 1
= 8² + 1
= 64 + 1
= 65
∴ The required Pythagorean triplet is 16, 63 and 65.

(iv) Let 2n = 18
n = \(\frac{18}{2}\) = 9

n² – 1
= 9² – 1
= 81 – 1
= 80

n² + 1
= 9² + 1
= 81 + 1
= 82
∴ The required Pythagorean triplet is 18, 80 and 82.

These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.1

Question 1.
What will be the unit digit of the squares of the following numbers?
(i) 81
(ii) 272
(iii) 799
(iv) 3853
(v) 1234
(vi) 26387
(vii) 52698
(viii) 99880
(ix) 12796
(x) 55555
Solution:
(i) The unit digit of (81)² is 1 (1 × 1 = 1)
(ii) The unit digit of (272)² is 4 (2 × 2 = 4)
(iii) The unit digit of (799)² is 1 (9 × 9 = 81)
(iv) The unit digit of (3853)² is 9 (3 × 3 = 9)
(v) The unit digit of (1234)² is 6 (4 × 4 = 16)
(vi) The unit digit of (26387)² is 9 (7 × 7 = 49)
(vii) The unit digit of (52698)² is 4 (8 × 8 = 64)
(viii) The unit digit of (99880)² is 0 (0 × 0 = 0)
(ix) The unit digit of (12796)² is 6 (6 × 6 = 36)
(x) The unit digit of (55555)² is 5 (5 × 5 = 25)

Question 2.
The following numbers are obviously not perfect squares. Give reason.
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050
Solution:
(i) 1057 is not a perfect square,
As the last digit is 7 [It is not one of 0, 1, 4, 5, 6, and 9].

(ii) 23453 is not a perfect square,
As the last digit is 3 [It is not one of 0, 1, 4, 5, 6, and 9].

(iii) 7928 is not a perfect square,
As the last digit is 8 [It is not one of 0, 1, 4, 5, 6, and 9].

(iv) 222222 is not a perfect square,
As the last digit is 2 [It is not one of 0, 1, 4, 5, 6, and 9].

(v) 64000 is not a perfect square,
As the number of zeros is odd.

(vi) 89722 is not a perfect square,
As the last digit is 2 [It is not one of 0, 1, 4, 5, 6, and 9].

(vii) 222000 is not a perfect square,
As the number of zeros is odd.

(viii) 505050 is not a perfect square,
As the number of zeros is odd.

Question 3.
The squares of which of the following would be odd numbers?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Note: The square of an odd natural number is always odd and that of an even number always an even number.
Solution:
(i) The square of 431 is an odd number.
(ii) The square of2826 is an even number.
(iii) The square of 7779 is an odd number.
(iv) The square of82004 is an even number.

Question 4.
Observe the following pattern and find the missing digits.
11² = 121
101² = 10201
1001² = 1002001
100001² = 1………2……..1
10000001² = …………….
Solution:
By observing the above pattern, we get
(i) 100001² = 10000200001
(ii) 10000001² = 100000020000001

Question 5.
Observe the following pattern and supply the missing number.
(a) 11² = 121
101² = 10201
10101² = 102030201
1010101² = …………….
………..² = 1020304030201
Solution:
By observing the above pattern, we get
(I) (1010101)² = 1020304030201
(ii) 10203040504030201 = (101010101)²

Question 6.
Using the given pattern. Find the missing numbers.
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + ……² = 21²
5² + …..² + 30² = 31²
6² + 7² + ……² = ……..²
Solution:
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² + 42² = 43²
Note: To find pattern
The third number is related to the first and second numbers. How?
The fourth number is related to the third number. How?

Question 7.
Without adding, find the sum
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
(i) Sum of the first 5 odd numbers = 5² = 25
(ii) Sum of the first 10 odd numbers = 10² = 100
(iii) Sum of the first 12 odd numbers = 12² = 144

Question 8.
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as to the sum of 11 odd numbers.
Solution:
(i) 7² = 49
1 + 3 + 5 + 7 + 9 + 11 + 13 (First 7 odd numbers)
(ii) 11² = 121
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (First 11 odd numbers)

Question 9.
How many numbers lie between squares of the following numbers?
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100
Solution:
Note: Since between n and (n + 1) there are 2n non-square numbers
(i) Between 12² and 13², there are 2 × 12 numbers i.e. 24 numbers
(ii) Between 25² and 26², there are 2 × 25 numbers i.e. 50 numbers
(iii) Between 99² and 100², there are 2 × 99 numbers i.e. 198 numbers
Note: For any number ending with 5, the square is a (a + 1) × 100 + 25
(i) 35²
= 3(3 + 1) × 100 + 25
= 1200 + 25
= 1225

(ii) 55²
= 5(5 + 1) × 100 + 25
= 3000 + 25
= 3025

(iii) 125²
= 12(12 + 1) × 100 + 25
= 12 × 13 × 100 + 25
= 15600 + 25
= 15625

These NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Exercise 5.1

Question 1.
Find the complement of each of the following angles.

Answer:
(i) Complement of
20° = 90° – 20° = 70°

(ii) Complement of
63° = 90° – 63° = 27°

(iii) Complement of 5 7° = 90° – 57°
= 33°

Quesrion 2.
Find the supplement of each of the following angles:


Answer:
(i) Supplement of
105°= 180°- 105° = 75°

(ii) Supplement of
87° = 180° – 87° = 93°

(iii) Supplement of
154° = 180° – 154° = 26°

Question 3.
Identify which of the following pairs of angles are complementary and which are supplementary.
(i) 65°, 115°
(ii) 63°, 27°
(iii) 112°, 68°
(iv) 130°, 50°
(v) 45°, 45°
(vi) 80°, 10°
Answer:
(i) 65°+ 115°= 180°
65 and 115° are supplementary angles.

(ii) 63°+ 27° = 90°
63 and 27° are complementary angles.

(iii) 112°+ 68°= 180°
112° and 68° are supplementary angles.

(iv) 130° + 50° = 180°
130° and 50° are supplementary angles.

(v) 45° + 45° = 90°
45° and 45° are complementary angles.

(vi) 80° + 10° = 90°
80° and 10° are complementary angles.

Question 4.
Find the angle which is equal to its complement.
Answer:
Let the required angle be x.
Given that the angle is equal to its complement
x = 90° – x
x + x = 90°
2x = 90°
x = \(\frac{90^{\circ}}{2}\) = 45°
Thus, 45° is equal to its complement.

Question 5.
Find the angle which is equal to its supplement.
Answer:
Let the required angle be ‘m’ and supplement of m = 180° – m
∴ m = 180° – m
(∵ m is equal to its supplement)
m + m = 180°
2m = 180°
m = \(\frac{180}{2}\) =90°
Thus, 90° is equal to its supplement.

Question 6.
In the given figure, ∠1 and ∠2 are supplementary angles.
If ∠1 is decreased, what changes should take place in ∠2 so that both the angles still remain supplementary.

Answer:
In case ∠1 is decreased, the same amount of degree measure is added to ∠2.
i.e. ∠2 be increased by same amount of degree measure.

Question 7.
Can two angles be supplementary if both of them are:
(i) acute?
(ii) obtuse?
(iii) right?
Answer:
(i) Sum of two acute angle is always less than 180°
∴ Two acute angles cannot be supplementary.

(ii) Sum of two obtuse angles is always more than 180°.
∴ Two obtuse angles cannot be supplementary.

(iii) Sum of two right angles =180°
∴ Two right angles can be supplementary.

Question 8.
An angle is greater than 45°. Is its complementary angle greater than 45° or equal to 45° or less than 45°?
Answer:
Complement of an angle (greater than 45°) is less than 45°.

Question 9.
In the figure given below:
(i) Is∠1 adjacent to ∠2?
(ii) Is ∠AOC adjacent to ∠AOE?
(iii) Do ∠COE and ∠EOD form a linear pair?
(iv) Are ∠BOD and ∠DOA supplementary?
(v) Is ∠1 vertically opposite to ∠4?
(vi) What is the vertically opposite angle of ∠5?

Answer:
(i) Yes, ∠1 and ∠2 are adjacent angles because both the angles have common arm OC and common vertex O.
(ii) No, ∠AOC is not adjacent to ∠AOE because ∠AOC is part of ∠AOE.
(iii) Yes, ∠COE and ∠EOD form a linear pair because COD is a straight line.
(iv) Yes, ∠BOD and ∠DOA are supplementary because ∠BOD + ∠DOA = 180°.
(v) Yes, because AB and CD intersect each other.
(vi) The vertically opposite angle of ∠5 is ∠BOC.

Q1uestion 10.
Indicate which pairs of angles are:
(i) Vertically opposite angles.
(ii) Linear pairs

Answer:
(i) Vertically opposite angles.
In the given figure, the following pairs are vertically opposite angles.
∠1 and ∠4
∠5 and (∠2 + ∠3)

(ii) Linear pairs:
∠4 and ∠5 form a linear pair.
∠1 and ∠5 form a linear pair.
∠1 and (∠3 + ∠2) form a linear pair.
∠4 and (∠3 + ∠2) form a linear pair.

Question 11.
In the following figure, is ∠1 adjacent to ∠2? Give reasons.

Answer:
∠1 and ∠2 are not adjacent angles because they do not have a common vertex.

Question 12.
Find the values of the angles x, y, and z in each of the following:

Answer:
(i) Since x and 55° are vertically opposite angles
x = 55°
Also, 55° + y = 180°
(Linear pair)
y = 180°- 55°
y = 125°
since y and z are vertically opposite angles
y = z = 125°
Thus x = 55°, y = 125° and z = 125°

(ii) since 40° and z are vertically
opposite angles
z = 40°
Again y and 40° form a Linear pair
y + 40° = 180°
y = 180°-40°
= 140°
y and (x + 25) are vertically opposite angles
(x + 25°) = y = 140°
∴ x + 25° = 140°
x = 140° -25°= 115°
Thus, x = 115°’ y = 140° and z = 40°

Question 13.
Fill in the blanks:
Answer:
(i) If two angles are complementary, then the sum of their measures is 90°.
(ii) If two angles are supplementary, then the sum of their measures is 180°.
(iii) Two angles forming a linear pair are supplementary.
(iv) If two adjacent angles are supplementary, they form a linear pair.
(v) If two lines intersect at a point, then the vertically opposite angles are always Equal.
(vi) If two lines intersect at a point, and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are obtuse angles.

Question 14.
In the adjoining figure, name the following pairs of angles
(i) Obtuse vertically opposite angles
(ii) Adjacent complementary angles
(iii) Equal supplementary angles
(iv) Unequal supplementary angles
(v) Adjacent angles that do not form a linear pair

Answer:
(i) ∠BOC and ∠AOD are obtuse vertically opposite angles.
(ii) ∠AOB and ∠AOE are adjacent complementary angles.
(iii) ∠BOE and ∠EOD are equal supplementary angles.
(iv) ∠AOE and ∠EOC are unequal supplementary angles.
(v) (a) ∠BOA and ∠AOE
(b) ∠AOE and ∠EOD
(c) ∠EOD and ∠COD

These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.3

Question 1.
Find the circumference of the circles with the following radius: (Take π = \(\frac{22}{7}\))
(a) 14 cm
(b) 28 mm
(c) 21 cm
Answer:
(a) Radius of the circle (r) = 14 cm
Circumference of the circle = 2πr units
= 2 × \(\frac{22}{7}\) x 14 cm
= 2 × 22 × 2 cm
= 88 cm

(b) Radius of the circle = 28 mm.
.’. Circumference of the circle = 2πr units
= 2 × \(\frac{22}{7}\) × 28 mm
= 2 × 22 × 4 mm
= 176 mm

(c) Radius of the circle = 21 cm
Circumference of the circle = 2πr 22
= 2 × \(\frac{22}{7}\) x 21
= 2 × 22 × 3 cm
= 132 cm

Question 2.
Find the area of the following circles, given that:
(a) Radius = 14 mm (π = \(\frac{22}{7}\))
(b) Diameter = 49 m
(c) Radius = 5 cm
Answer:
(a) Radius of the circle = 14 mm
Area of the circle = πr² sq.m
= \(\frac{22}{7}\) × 14²
= \(\frac{22}{7}\) × 196
= 22 × 28
= 616 mm²

(b) Diameter of the circle = 49 m.
Radius of the circle = \(\frac{49}{2}\) m
= 24.5 m
Area of the circle = πr²
= \(\frac{22}{7}\) × (24.5)²
= \(\frac{22}{7}\) × 600.25
= 22 × 85.75
= 1886.5 m²
Area of the circle = πr² sq units

(c) Radius of the circle = 5 cm
Area of the circle = πr² sq.unit
= \(\frac{22}{7}\) × 5 × 5 cm²
= \(\frac{22 \times 25}{7}\) cm²
= \(\frac{550}{7}\) cm² or 78.57 cm².

Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = \(\frac{22}{7}\))
Answer:
Circumference of a circle = 154 m
2πr = 154
2 × \(\frac{22}{7}\) × r = 154

Area of the circle = πr² sq.m
= \(\frac{22}{7}\) × (24.5)²
= \(\frac{22}{7}\) × 600.25
= 22 × 85.75 m²
= 1886.5 m²

Question 4.
A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also, find the cost of the rope, if it cost ₹ 4 per meter. (Take π = \(\frac{22}{7}\))

Answer:
Diameter of the circular garden = 21 m
radius of the circular garden = \(\frac{21}{2}\)
Circumference of the garden = 2πr units
= 2 × \(\frac{22}{7}\) × \(\frac{21}{2}\) m = 22 x 3m = 66m
Length of the rope required for one round fence = 66 m
Length of the rope required for two round fence = 66 m × 2 m = 132 m
Cost of rope per meter = ₹ 4
Total cost of the rope = ₹132 × 4 = ₹ 528

Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

Answer:
Radius of outer circle (R) = 4 cm
Radius of inner circle (r) = 3 cm
= Area of the outer circle – Area of the inner circle
Area of the remaining sheet = n (R² – r²)
= 3.14 (4² – 3²)
= 3.14 × (16-9)
= 3.14 x 7 = 21.98 cm²
So, the area of the remaining sheet is 21.98 cm².

Question 6.
Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ 15. (Take π = 3.14)
Answer:
Diameter of the table cover = 1.5 m
Radius of the table cover = \(\frac { 1.5 }{ 2 }\) m
Circumference of the table cover = 2πr units
= 2 x 3.14 x \(\frac { 1.5 }{ 2 }\) = 3.14 x 1.5 m = 4.71 m
Cost of lace for one metre = ₹15
Cost of lace for table cover = ₹ 4.71 x 15
= ₹ 70.71

Question 7.
Find the perimeter of the given figure, which is a semicircle including its diameter.

Answer:
Diameter of the semicircle = 10 cm
Radius of the semicircle = \(\frac { 10 }{ 2 }\) = 5 cm
Circumference of the semicircle = \(\frac { 1 }{ 2 }\) x 2πr unit
= πr = \(\frac { 22 }{ 7 }\) x 5
= 15.71 cm
∴ Perimeter of the semicircle
= 15.71 + 10 cm
= 25.71 cm

Question 8.
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15/m². (Take π = 3.14)
Answer:
Diameter of the table-top = 1.6 m
Radius of the table-top = \(\frac { 1.6 }{ 2 }\) = 0.8m
Area of the table-top = πr² sq. m
= 3.14 × 0.8 × 0.8 m² = 2.0096 m²
Rate of Polishing = ₹ 15 per m²
Cost of polishing the table-top = ₹ 2.0096 × 15 =₹ 30.14 (approx.)

Question 9.
Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also, find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square?
(Take π = \(\frac { 22 }{ 7 }\))
Answer:
Length of the wire = 44 cm
Let the radius of the circle be r.
Circumference of the circle = 44 cm
2πr = 44
2 × \(\frac { 22 }{ 7 }\) × r = 44
r = \(\frac{44 \times 7}{2 \times 22}\) cm =7 cm
Area of the circle = πr² sq.m
= \(\frac { 22 }{ 7 }\) × 7 × 7 cm²
= 154 cm

Since, the wire is rebent to form a square. Perimeter of the square = Length of the wire
4a = 44
[Perimeter of a square = 4a] 44
a = \(\frac { 44 }{ 6 }\) = 11 cm 6
Side of a square =11 cm
Area of the square = side × side
= 11 × 11 cm² = 121 cm²
154 cm² > 121 cm²
Area of the circle > Area of the square
∴ The circle encloses greater area.

Question 10.
From a circular card sheet of radius 14cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed, (as shown in the given figure). Find the area of the remaining 22
sheet. (Take π = \(\frac { 22 }{ 7 }\))

Answer:
Radius of the circular card sheet =14 cm
Area of the sheet = πr² sq m
= \(\frac { 22 }{ 7 }\) × 14 × 14 cm²
= 22 × 2 × 14 cm2 = 616 cm²
Radius of a small circle = 3.5 cm
Area of 2 small circles = 2 × πr² sq m
= 2 × \(\frac { 22 }{ 7 }\) × 3.5 × 3.5 cm²
= 2 × 22 × 0.5 × 3.5 cm = 77 cm²
Length of a small rectangle = 3 cm
Breadth of a small rectangle = 1 cm.
Area of small rectangle = l × b sq.m
= 3 × 1 cm² = 3 cm²
Area of the remaining sheet = Area of the circular sheet – (Area of two small circles + Area of rectangle)
= 616 cm²– (77 + 3)cm²
= 616 cm² – 80 cm² = 536 cm²
∴ Required area of the remaining sheet = 536 cm²

Question 11.
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)
Answer:
Side of the square = 6 cm
Area of the square = side × side
= 6 cm × 6 cm = 36 cm²

Radius of the circle cut out from the sheet = 2 cm
Area of the circle = πr² sq.m
= 3.14 × 2 × 2 cm² = 12.56 cm²
Area of the remaining sheet
= 36 cm² – 12.56 cm² = 23.44 cm²

Question 12.
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take n = 3.14)
Answer:
Let the radius of the circle be ‘r’
Circumference of a circle = 31.4 cm
2πr² = 31.4
2 × 3.14 × r = 31.4
r = \(\frac{31.4}{2 \times 3.14}\)
= \(\frac{314 \times 10}{2 \times 314}\)
= \(\frac{10}{2}\)
= 5 cm
Area of the circle = 2πr² sq. units = 3.14 × 5 × 5 cm² = 78.5 cm²

Question 13.
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower
bed is 66 m. What is the area of this path? (π = 3.14)
Answer:
Diameter of the flower bed = 66 m
Radius of the flower bed = \(\frac { 66 }{ 2 }\) m
(r) = 33 m

width of the surrounding path = 4 m
Radius of the outer circle (R)
= 33 m + 4 m
(R) = 37 m
Area of the pathway = Area of the outer circle – Area of the inner circle
= πR² – πr²
= π(R² – r²) sq. units
= 3.14 (37² – 33² ) m²
= 3.14 (37 + 33) (37 – 33) m²
= 3.14 × 70 × 4 m²
= 3.14 × 280 m²
= 879.2 m²
Thus, the area of the path = 879.2 m²

Question 14.
A circular flower garden has an area of 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water
the entire garden? (Take π = 3.14)
Answer:
Let the radius of the garden be ‘r’
Area of the circular garden = 314 m²
πr² =314
3.14 × r² = 314
r² = \(\frac{314}{3.14}=\frac{314 \times 100}{314}\)
r² = 100
r²= 10²
r = 10 m
Radius of the area covered by the sprinkler = 12 m
Since 12 m > 10 m
The sprinkler covers an area beyond the garden.
Yes, the entire garden is covered by the sprinkler.

Question 15.
Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Answer:
Radius of the outer circle (R) = 19 m.
Circumference of the outer circle
= 2πr unit
= 2 × 3.14 × 19 m
= 119.32 m
Radius of the inner circle (r)
= 19- 10 = 9m
Circumference of the inner circle
= 2π = 2 × 3.14 × 9 m = 56.52 m

Question 16.
How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = \(\frac { 22 }{ 7 }\))
Answer:
Radius of a wheel = 28 cm
∴ Circumference of the wheel = 2πr units
= 2 × \(\frac { 22 }{ 7 }\) × 28
= 2 × 22 × 4 cm = 176 cm
Total distance covered
= 352 m
= 352 × 100 cm
= 35200 cm
Number of rotations
= \(\begin{aligned}
&\frac{\text { Total distance }}{\text { umference of the wheel }}\\&\text { Circum }
\end{aligned}\)
= \(\frac{35200}{176}\) = 200
Thus, the distance of 352 m will be covered in 200 rotations.

Question 17.
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take π = 3.14)
Answer:
Length of the minute hand =15 cm
radius of the circle (r) = 15 cm
(made by the tip of the minute hand)
Perimeter of the circle = 2πr units
= 2 × 3.14 × 15 cm = 94.2 cm
Distance covered by the minute hand = 94.2 cm

These NCERT Solutions for Class 8 Maths Chapter 5 Data Handling InText Questions Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling InText Questions

Try These (Page No. 71)

Question 1.
Draw an appropriate graph to represent the given information.

Solution:
(i) Note: A bar graph showing two sets of data simultaneously is called a double-bar graph. It is useful for the comparison of the data.

To represent the given data by a bar- graph, draw two axes perpendicular to each other. Now, represent ‘Months’ on OX and ‘Number of watches sold’ on OY. Erect recharge of the same width. The heights of the rectangles are proportional to the number of watches using a suitable scale:
Here, the scale is 1 cm = 500 watches
Since 500 watches = 1 cm
1000 watches = 2 cm
1500 watches = 3 cm
2000 watches = 4 cm
2500 watches = 5 cm

(ii)

(iii) Since, a comparison of two activities (walking and cycling) is to be represented, there a double-graph is drawn by taking the schools along the x-axis and the number of children on the y-axis, using a scale of 1 cm = 5 children.

To compare the percentage win in ODI achieved by various teams, we represent the data by a double-bar graph. We represent the teams along the x-axis and their percentage win along the y-axis, using the scale 1 cm = 5%.

Try These (Page No. 72)

Question 2.
A group of students was asked to say which animal they would like most to have as a pet. The results are given below:
dog, cat, cat, fish, cat, rabbit, dog, cat, rabbit, dog, cat, dog, dog, dog, cat, cow, fish, rabbit, dog, cat, dog, cat, cat, dog, rabbit, cat, fish, dog.
Make a frequency distribution table for the same.
Solution:
Using tally-marks, we have:

Try These (Page No. 73-74)

Question 3.
Study the following frequency distribution table and answer the questions given below.
Frequency Distribution of Daily Income of 550 workers of a factory
Table 5.3

(i) What is the size of the class intervals?
(ii) Which class has the highest frequency?
(iii) Which class has the lowest frequency?
(iv) What is the upper limit of the class interval 250 – 275?
(v) Which two classes have the same frequency?
Solution:
(i) Class size = (Upper class limit) – (Lower class limit)
= 125 – 100
= 25
(ii) The class 200 – 225 is having the highest frequency (which is 140).
(iii) The class 300 – 325 is having the lowest frequency (which is 20).
(iv) The upper limit of the class interval 250 – 275 is 275.
(v) The classes (150 – 175) and (225 – 250) are having the same frequency (which is 55).

Question 4.
Construct a frequency distribution table for the data on weights (in kg) of 20 students of a class using intervals 30 – 35, 35 – 40, and so on.
40, 38, 33, 48, 60, 53, 31, 46, 34, 36, 49, 41, 55, 49, 65, 42, 44, 47, 38, 39.
Solution:
Lowest observation = 31
Highest observation = 65
Class intervals: 30 – 35, 35 – 40, 40 – 45,…..
The frequency distribution table for the above data can be:

Try These (Page No. 75)

Question 5.
Observe the histogram (Fig 5.3) and answer the questions given below.

(i) What information is being given by the histogram?
(ii) Which group contains maximum girls?
(iii) How many girls have a height of 145 cm and more?
(iv) If we divide the girls into the following three categories, how many would there be in each?
150 cm and more – Group A
140 cm to less than 150 cm – Group B
Less than 140 cm – Group C
Solution:
(i) The above histogram represents the height (in cms) of girls of Class VIII.
(ii) The group 140 – 145 contains a maximum number of girls (which has as many as 7 girls).
(iii) 7 girls (= 4 + 2 + 1) have a height of 145 cm and more.
(iv) Number of girls in
Group A: 150 cm and more = 2 + 1 = 3 girls
Group B: 140 cm to less than 150 cm = 7 + 4 = 11 girls
Group C: Less than 140 cm = 1 + 2 + 3 = 6 girls.
Note: The broken line (~) is used along the horizontal line to indicate that we are not showing the numbers between 0 and 125.

Try These (Page No. 78)

Question 6.
Each of the following pie charts (Fig 5.5) gives you a different piece of information about your class.
Find the fraction of the circle representing each of this information.

Solution:
(i) Fraction of the circle representing the ‘girls’ 50% = \(\frac{50}{100}=\frac{1}{2}\)
Fraction of the circle representing the ‘boys’ 50% = \(\frac{50}{100}=\frac{1}{2}\)

(ii) Fraction of the circle representing ‘walk’ 40% = \(\frac{40}{100}=\frac{2}{5}\)
Fraction of the circle representing ‘but or car’ 40% = \(\frac{40}{100}=\frac{2}{5}\)
Fraction of the circle representing ‘cycle’ 20% = \(\frac{20}{100}=\frac{1}{5}\)

(iii) Fraction of the circle representing those who live mathematics = (100 – 15)%
= \(\frac{100-15}{100}\)
= \(\frac{85}{100}\)
= \(\frac{17}{20}\)
Fraction of the circle representing those who hate mathematics = 15%
= \(\frac{15}{100}\)
= \(\frac{3}{20}\)

Question 7.
Answer the following questions based on the pie chart given.
(i) Which type of programmes are viewed the most?
(ii) Which two types of programmes have a number of viewers equal to those watching sports channels?

Solution:
From the given pie chart, we have: Obviously,

(i) The entertainment programmes are viewed the most.
(ii) The news and informative programmes have an equal number of viewers.

Try These (Page No. 81)

Question 8.
Draw a pie chart of the data given below.
The time spent by a child during a day.
Sleep – 8 hours
School – 6 hours
Homework – 4 hours
Play – 4 hours
Others – 2 hours
Solution:
First, we find the central angle corresponding to the given activities.

Now, the required pie chart is given below:

Try These (Page No. 83)

Question 9.
If you try to start a scooter, what are the possible outcomes?
Solution:
It may start. It may not start.

Question 10.
When a die is thrown, what are the possible outcomes?
Solution:
The possible outcomes are 1, 2, 3, 4, 5, or 6.

Question 11.
When you spin the wheel shown, what are the possible outcomes? List them.
(Outcome here means the sector at which the pointer stops).

Solution:
The possible outcomes are A, B, or C.

Question 12.
You have a bag with five identical balls of different colours and you are to pull out (draw) a ball without looking at it; list the outcomes you would get.

Solution:
The possible outcomes are W, R, B, G, or Y.