Samacheer Class 7 Maths - Measurements Intext Questions

Given radius of the bangle = 2 inches
Diameter = 2 × radius = 2 × 2 = 4 inches

Exercise 2.2
Try These (Text book Page No. 33)

(i) Radius of the circle r = 4.2 cm
Area of the circle A = π r 2 sq.units
= \(\frac { 22 }{ 7 } \) × 4.2 × 4.2 cm 2 = 5.44 cm 2
(ii) Diameter of the circle d = 28 cm
radius r = \(\frac { d }{ 2 } \) = \(\frac { 28 }{ 2 } \) = 14 cm
Area of the circle A = π r 2 sq.units
= \(\frac { 22 }{ 7 } \) × 14 × 14 cm 2 = 616 cm 2

Exercise 2.3
Try These (Text book Page No. 35)

Radius of the outer circle R = 9 cm
Radius of the inner circle r = 6 cm
Width of the circular pathway = Radius of the outer circle
– Radius of the inner circle
= (9 – 6) cm = 3 cm
Width of the circular pathway = 3 cm

Given outer radius R = 16 cm
Area of the circular pathway = πR 2 = πr 2
Area of the circular pathway = 352 sq. cm
πR 2 – πr 2 = 352 cm 2
π(R 2 – r 2 ) = 352
16 2 – r 2 = \(\frac{352 \times 7}{22}\)
16 2 – r 2 = 16 × 7
16 2 – r 2 = 112
16 2 – 112 = r 2
r 2 = 256 – 112
r 2 = 144
r = 12 cm
Inner radius r = 12 cm

Area of the outer rectangle = 48 sq.cm
Area of the inner rectangle = 15 sq.cm
Area of the rectangular pathway = Area of the outer rectangle
– Area of the inner rectangle
= 48 – 15 = 33 cm 2
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(i) Diameter d = 70 cm
Circumference C = π d units = \(\frac { 22 }{ 7 } \) × 70 = 22 × 10 = 220 cm
(ii) Diameter d = 56 m
Circumference = π d units
= \(\frac { 22 }{ 7 } \) × 56 = 22 × 8 = 176 m
(iii) Diameter d = 28 mm
Circumference C = π d units = \(\frac { 22 }{ 7 } \) × 28 = 22 × 4 = 88 mm

Radius r = 49 cm
Circumference C = 2 π r units = 2 × \(\frac { 22 }{ 7 } \) × 49 = 2 × 22 × 7
= 44 × 7 = 308 cm
(ii) Radius r = 91 mm
Circumference C = 2 π r units
= 2 × \(\frac { 22 }{ 7 } \) × 91 = 2 × 22 × 13 = 44 × 13 = 572 mm

Given the diameter d = 4.2 m
Circumference C = π d units = \(\frac { 22 }{ 7 } \) × 4.2 m = 22 × 0.6 = 13.2 m

Diameter of the bullock cart wheel d= 1.4 m
Distance covered in 1 rotation = Its circumference
= π d units = \(\frac { 22 }{ 7 } \) × 1 .4 m = 22 × 0.2 = 4.4 m
Distance covered in one rotation = 4.4 m
Distance covered in 150 rotations = 4.4 × 150 = 660.0
Distance covered in 150 rotations = 660 m

Diameter of the ground d = 350 m
Distance covered in 1 revolution = Circumference of the circle
= π d units = \(\frac { 22 }{ 7 } \) × 350 m = 22 × 50 = 1100 m
Distance covered in 1 rotation = 1100 m
Distance covered in 4 revolutions = 1100 × 4 = 4400 m

Length of the wire = 1320 cm
Radius of each circular frame = 7cm
Circumference of the frame 2 π r units = 2 × \(\frac { 22 }{ 7 } \) × 7 cm = 2 × 22 = 44 cm

30 frames can be made.

Radius of the garden r = 63 m
Circumference of the garden = 2 π r units = 2 × \(\frac { 22 }{ 7 } \) × 63 m = 2 × 22 × 9 = 396 m
Cost of fencing 1 meter = ₹ 150
Cost of fencing 396 meter = ₹ 150 × 396 = ₹ 59,400
∴ Cost of fencing the garden = ₹ 59,400

Diameter of the dinig table (d) = 105 cm
∴ Radius r = \(\frac { d }{ 2 } \) = \(\frac { 105 }{ 2 } \) cm
Area of the circle = π r 2 = \(\frac { 22 }{ 7 } \) × \(\frac { 105 }{ 2 } \) × \(\frac { 105 }{ 2 } \) = 8662.5 sq.cm
Area of the dinning table = 8662.5 cm 2

Radius of the shotput ring r = \(\frac { d }{ 2 } \) = \(\frac { 2.135 }{ 2 } \) m
Area of the circle = π r 2
= \(\frac { 22 }{ 7 } \) × \(\frac { 2.135 }{ 2 } \) × \(\frac { 2.135 }{ 2 } \)
= \(\frac { 25.07 }{ 7 } \) = 3.581 m 2
∴ Area of the shotput ring = 3.581 m 2

Area of the Circle = π r 2 sq.units
Area of the circular portion watered = 1386 cm 2
π r 2 = 1386
\(\frac { 22 }{ 7 } \) × r 2 = 1386
r 2 = 1386 × \(\frac { 7 }{ 22 } \) = 63 × 7 = 9 × 7 × 7
r 2 = 3 2 × 7 2
r = 3 × 7
Radius (r) = 21 cm
Diameter (d) = 2 r = 2 × 21 cm
Diameter (d) = 42 cm

Circumference of a Circle = 2 π r units
Given circumference of a circular park = 352 m
2 π r = 352
2 × \(\frac { 22 }{ 7 } \) × r = 352
r = 352 × \(\frac { 7 }{ 22 } \) × \(\frac { 1 }{ 2 } \) = 56 m
Area of the park = π r 2 = \(\frac { 22 }{ 7 } \) × 56 × 56 sq.units
= 22 × 8 × 56 = 9856 m 2
∴ Area of the Circular park = 9856 m 2

Length of the rope = 4.9 m
Area that the sheep can graze = Area of circle with radius 4.9m
Area of the circle = π r 2 sq.units
= \(\frac { 22 }{ 7 } \) × 4.9 × 4.9 = 22 × 0.7 × 4.9 = 75.46
∴ Area that the sheep can graze = 75.46 m 2

If the bull is tethered by a rope then the area it can graze is a circular area of radius
= length of the rope
Area of the circle = 2464 m 2
π r 2 = 2464 m 2
\(\frac { 22 }{ 7 } \) × r 2 = 2464
r 2 = 2464 × \(\frac { 7 }{ 22 } \) = 122 × 7 = 16 × 7 × 7
r 2 = 42 × 72
r = 4 × 7 = 28 m
length of the rope r = 28 m

Radius of the round carpet = 63 cm
Area covered by the round carpet = πr 2 sq units
A = \(\frac { 22 }{ 7 } \) × 63 × 63 = 22 × 9 × 63 = 12474 cm 2
Area covered by the round carpet = 12,474 cm 2

Diamter of the circular garden d = 49 m
Radius r = \(\frac { d }{ 2 } \) = \(\frac { 49 }{ 2 } \) m
Area of the circular garden = πr 2 sq units
= \(\frac { 22 }{ 7 } \) × \(\frac { 49 }{ 2 } \) × \(\frac { 49 }{ 2 } \) m 2 = 1,886.5 m 2
Cost of levelling a m 2 area = ₹ 150
∴ Cost of levelling 1886.5 m 2 = ₹ 150 × 1886.5 = ₹ 2,82,975
Cost of levelling the flower garden = ₹ 2,82,975

Radius of the circular swimming pool r = 7 m
Area of the circular swimming pool A = πr 2 sq. units
= \(\frac { 22 }{ 7 } \) × 7 × 7 m 2 = 154 m 2
Cost of cementing a m 2 floor = ₹ 18.
Cost of cementing 154 m 2 floor = ₹ 18 × 154 = ₹ 2,772

Radius of the outer circle R = 32 cm
Radius of the inner circle r = 18 cm
Area of the circular pathway = π (R 2 – r 2 ) sq. units = \(\frac { 22 }{ 7 } \) (32 2 – 18 2 ) cm 2
= \(\frac { 22 }{ 7 } \) × (32 + 18) × (32 – 18) cm 2
= \(\frac { 22 }{ 7 } \) × 50 × 14 cm 2 = 2,200 cm 2
Area of the circular pathway = 2,200 cm 2

Radius of the circular lawn r = 28 m
Radius of the lawn with path = 28 + 7 m = 35 m
Area of the circular path = π (R 2 – r 2 ) sq. units
Area of the path = \(\frac { 22 }{ 7 } \) (35 2 – 28 2 ) m 2 = \(\frac { 22 }{ 7 } \) × (35 + 28) (35 – 28) m 2
= \(\frac { 22 }{ 7 } \) × 63 × 7 m 2 = 1386 m 2
Area of the path = 1386 m 2

Radius of the circular hall R = 120 cm
Radius of the circular carpet r = 106 cm
Area of the hall uncovered = Area of the hall – Area of the carpet
= π (R 2 – r 2 ) cm 2
= \(\frac { 22 }{ 7 } \) × (120 2 – 106 2 ) cm 2
= \(\frac { 22 }{ 7 } \) × (120 + 106) × (120 – 106) cm 2
= \(\frac { 22 }{ 7 } \) × 226 × 14 cm 2 = 9,944 cm 2
Area of the hall uncovered = 9, 944 cm 2

Radius of the ground R = 103 m
Width of a track W = 3 m
Width of 4 tracks = 4 × 3 = 12 m
Radius of the ground without track
r = (103 – 12)m
r = 91 m
Area of 4 tracks = Area of the ground
– Area of the ground without crack
= πR 2 – πr 2 sq.units
= π(R 2 – r 2 ) sq.units
= \(\frac { 22 }{ 7 } \) [103 2 – 91 2 ]
= \(\frac { 22 }{ 7 } \) [103 + 91] [103 – 91]m 2
= \(\frac { 22 }{ 7 } \) × 194 × 12 = \(\frac { 51216 }{ 7 } \) = 7316.57 m 2
∴ Area of 4 tracks = 7316.57 m 2
Cost of constructing 7316.57 m 2 = ₹ 50
∴ Cost of constructing 7316.57 m 2 = ₹ 50 × 7316.57 = ₹ 3,65,828,57
Cost of constructing the track ₹ 3,65,828,57

Area of the rectangle = (Lenght × Breadth) sq. units
Area of the outer rectangle = (L × B) sq. units
Length of the outer rectangle L = 80 m
Breadth of the outer rectangle B = 50 m
Length of the inner rectangle l = 70 m
Breadth of the inner rectangle b = 40 m
Area of the outer rectangle = 80 × 50 m 2 = 4000 m 2
Area of the inner rectangle = l × b sq. unit = 70 × 40 m 2 = 2800 m 2
Area of the pathway = Area of the outer rectangle
– Area of the inner rectangle
= 4000 – 2800 m 2 = 1200 m 2
Area of the pathway = 1200 m 2

Area of the rectangular garden L × B = 11 m × 8 m = 88 m 2
Length of the inner rectangle L = L – 2 W = 11 – 2(2) = 11 – 4 = 7 m
Breadth of the inner rectangle b = B – 2W = 8 – 2(2) = 8 – 4 = 4 m
Area of the inner rectangle = l × b sq. units = 7 × 4 m 2 = 28 m 2
Area of the path = Area of the outer rectangular garden
– Area of the inner rectangle
= 88 m 2 – 28m 2 = 60 m 2
Area of the path = 60 m 2

Length of the ceiling L = 18 m
Breadth of the ceiling B = 7 m
Area of the ceiling = L × B sq. units = 18 × 7 m 2 = 126 m 2
Width of the boarder W = 10 cm = \(\frac { 10 }{ 100 } \) m = 0.1 m
Length of the ceiling without border = L – 2W = 18 – 2(0.1) m
= 18 – 0.2 m = 17.8 m
Breadth of the ceiling without border = B – 2W = 7 – 2 (0.1) m
= 7 – 0.2 m = 6.8 m
Area of the ceiling without border = l × b sq.units
= 17.8 × 6.8 m 2 = 121.04 m 2
∴ Area of the border = Area of the ceiling
– Area of the ceiling without border
= 126 – 121.04 m 2 = 4.96 m 2
Area of the border = 4.96 m 2

Length of the field L = 24 m
Width (Breadth) of the field B = 15 m
(i) Area of the field = L × B sq. units = 24 × 15 m 2 = 360 m 2
(ii) Width of the canal (W) = 1 m
Length of the field without canal (l) = L – 2(W) = 24 – 2(1) m
= 24 – 2 m = 22 m
Width of the field without canal (b) = B – 2W = 15 – 2(1) m
= 15 – 2 m = 13 m
Area of the field without canal = l × b sq. units = 22 × 13 m 2 = 286 m 2
Area of the canal = 360 – 286 = 74 m 2
Cost of constructing 1 m 2 canal = ₹ 12
Cost of the constructing 74 m 2 canal = ₹ 12 × 74 = ₹ 888

Objective Type Question

Cost of fencing the circumference = ₹ 2112
Cost of fencing one meter = ₹ 8
∴ Circumference of the circle = \(\frac { 2112 }{ 8 } \) = 264 m
πd = 264 m
\(\frac { 22 }{ 7 } \) × d = 264
d = \(\frac{264 \times 7}{22}\) = 12 × 7 m = 84 m
∴ Diameter of the race cource = 84 m

Length of the rectangular ground l = 120 m
Breadth b = 90 m
Length of the path W 1 = 2m
Length of the path W 2 = 1m
Length of the ground with path L = 1 + 2 (W 2 ) = 120 + 2(1) m
= 120 + 2 = 122 m
Breadth of the ground with path B = l + 2(W 1 ) units
= 90 + 2(2) m = 90 + 4 m = 94 m
∴ Area of the path = (L × B) – (1 × b) sq. units
= (122 × 94) – (122 × 94) m 2 = 668 m 2
∴ Area of the path = 668 m 2

Cost of decorating the circumference = ₹ 16,940
Cost of decorating per meter = ₹ 55
∴ Length of the circumference = \(\frac { 16940 }{ 55 } \) m = 308 m
Circumference of the circular lawn = 308 m
2 × πr = 308 m
2 × \(\frac { 22 }{ 7 } \) × r = 308 m
r = \(\frac{308 \times 7}{2 \times 22}\)
r = 49 m
Radius of the lawn = 49 m

Given radius of a circle r = 3 cm
Diameter of the circle = 2r = 2 × 3 = 6 cm
Breadth of the rectangle = Diamter of the circle
B = 6cm
Length of the rectangle L = 4 × diameter of a circle
L = 4 × 6
L = 24cm
(i) Area of the rectangle = L × B sq. units
= 24 × 6 cm 2
Area of the rectangle = 144 cm 2
(ii) Area of the circle = πr 2 sq. units
= \(\frac { 22 }{ 7 } \) × 3 × 3 cm 2
= \(\frac { 198 }{ 7 } \) cm 2
= 28.28 cm 2
(iii) Area of the shaded area = Area of the rectangle – Area of the 4 circles
= 144 – (4 × \(\frac { 198 }{ 7 } \)) cm 2 = 144 – \(\frac { 792 }{ 7 } \) cm 2
= 144 – 113.14 cm 2 = 30.85 cm 2

Challenge Problems

Outer circumference of the circular lawn = 88 cm
2πR = 88 cm
Inner circumference of the lawn 2πr = 44 cm
2πR – 2πr = 88 – 44
2 × \(\frac { 22 }{ 7 } \) (R – r) = 44
(R – r) = \(\frac{44 \times 7}{2 \times 22}\)
Outer radius – Inner radius = 7 cm
∴ Width of the lawn = 7 cm
Also 2πR + 2πr = 88 + 44
2π (R + r) = 132
π (R + r) = \(\frac { 132 }{ 2 } \) = 66 cm
Area of the path = πR 2 – πr 2 sq. units
= π (R + r) (R – r) = 66 × 7
Area of the path = 462cm 2

Length of the field l = 76 m
Breadth of the field b = 60m
Area of the field A = l × b sq. units = 76 × 60 m 2
Area of the field A = 4560 m 2
Length of the rope = 35m
Radius of the land that the cow can graze = 35m
Area of the land tha the cow can graze = circle of radius 35 m = πr 2 sq.units
π × 35 × 35 m 2 = \(\frac { 22 }{ 7 } \) × 35 × 35 m 2
= 3850 m 2
Area of the land the cow cannot graze = Area of the field – Area that the cow can graze
= 4560 – 3860 m 2 = 710 m 2
Area of the land that the cow cannot graze = 710 m 2

Let the length of the rectangular field = ‘L’ m
Breadth of the rectangular field = = ‘B’ m
Area of the rectangular field = (L × B) m 2
Also given length = 3 × Breadth
L = 3B
Width of the path (W) = 5m
Lenth of the inner rectangle = L – 2W = l – 2(5)
= 3B – 10m
Breadth of the inner rectangle = B – 2W
= B – 2(5)
= B – 10 m
Area of the inner rectangle = (3B – 10) (B – 10)
= 3B 2 – 10B – 30B + 100
Area of the path = Area of outer rectangle
– Area of inner rectangle
= (L × B) – (3B 2 – 10B – 30B + 100)
3B × B – (3B 2 – 40B + 100)
= 3B 2 – 3B 2 + 40B – 100
Area of the path = 40B – 100
Given area of the path = 500 m 2
40B – 100 = 500
40B = 500 + 100 = 600
B = \(\frac { 600 }{ 40 } \)
B = 15m
Length of the field = 45 m; Breadth of the field = 15 m

Area of the outer circle = 1386 m 2
πR 2 = 1386m 2
Area of the inner circle = 616 m 2
πr 2 = 616m 2
Area of the path = Area of outer circle – Area of the inner circle
1386 m 2 – 616 m 2
Area of the path = 770m 2
Also πR 2 = 1386
R 2 = \(\frac{1386 \times 7}{22}\)
R 2 = 63 × 7
R 2 = 9 × 7 × 7
R 2 = 32 × 72
R = 3 × 7
Outer Radius R = 21 m
Again πr 2 = 616
\(\frac { 22 }{ 7 } \) × r 2 = 616
r 2 = 28 × 7
r 2 = 4 × 7 × 7
r 2 = 22 × 72
r = 2 × 7
Inner radius r = 14m
Width of the path = Outer radius – Inner radius = 21 – 14
Width of the path = 7m

Length of the rope = 45 m = Radius of the inner circle
∴ Area of the circular area that the goat graze = πr 2 sq. units
= \(\frac { 22 }{ 7 } \) × 45 × 45 m 2 = 6364.28 m 2
Radius of the gross land = 52 m
Area of the grass land = \(\frac { 22 }{ 7 } \) × 52 × 52 = 8,498.28 m 2
Area that the goat cannot graze
= Area of the outer circle – Area of the inner circle
= 8498.28 – 6364.28 = 2134 m 2
Area of the goat cannot grass = 2134 m 2

Area of the outer rectangular cardboard
= L × B sq.units = 30 × 20 cm 2 = 600 cm 2
Width of the stip = 4 cm
Length of the inner rectangle = L – 2W
l = 30 – 2(4) = 30 – 8
l = 22cm
Breadth of the inner rectangle B = 2W = 20 – 2(4) = 20 – 8
b = 12cm
Area of the inner rectangle = l × b sq.units = 22 × 12 cm 2 = 264 cm 2
Area of the remaining cardboard = 264 cm 2
Area of the removed portion = Area of outer rectangle
– Area of the inner rectangle
= 600 – 264 cm 2
Area of the removed portion = 336 cm 2

Length of the rectangular field L = 20 m
Breadth B = 15m
Area = L × B
20 × 15 m 2
Area of outer rectangle = 300 m 2

Area of inner small rectangle = \(\frac { 19 }{ 2 } \) × \(\frac { 13 }{ 2 } \) = 61.75 cm 2
(i) Area of the path = Area of the outer rectangle
– Area of 4 inner small rectangles
= 300 – 4(61.75) = 300 – 247 = 53 m 2
Area of the paths = 53 m 2
(ii) Area of the remaining portion of the field
= Area of the outer rectangle – Area of the paths
= 300 – 53 m 2 = 247 m 2
Area of the remaining portion = 247 m 2
(iii) Cost of constructing 1 m 2 road = ₹10
∴ Cost of constructing 53 m 2 road = ₹10 × 53 = ₹530
∴ Cost of constructing road = ₹530
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