Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.3

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.3

Question 1.
Find the area of a circular pathway whose outer radius is 32 cm and inner radius is 18 cm.
Solution:
Radius of the outer circle R = 32 cm
Radius of the inner circle r = 18 cm
Area of the circular pathway = π (R² – r²) sq. units = \(\frac { 22 }{ 7 } \) (32² – 18²) cm²
= \(\frac { 22 }{ 7 } \) × (32 + 18) × (32 – 18) cm²
= \(\frac { 22 }{ 7 } \) × 50 × 14 cm² = 2,200 cm²
Area of the circular pathway = 2,200 cm²

Question 2.
There is a circular lawn of radius 28 m. A path of 7 m width is laid around the lawn. What will be the area of the path?
Solution:
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 (R² – r²) sq. units
Area of the path = \(\frac { 22 }{ 7 } \) (35² – 28²) m² = \(\frac { 22 }{ 7 } \) × (35 + 28) (35 – 28) m²
= \(\frac { 22 }{ 7 } \) × 63 × 7 m² = 1386 m²
Area of the path = 1386 m²

Question 3.
A circular carpet whose radius is 106 cm is laid on a circular hall of radius 120 cm. Find the area of the hall uncovered by the carpet.
Solution:
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 (R² – r²) cm²
= \(\frac { 22 }{ 7 } \) × (120² – 106²) cm²
= \(\frac { 22 }{ 7 } \) × (120 + 106) × (120 – 106) cm²
= \(\frac { 22 }{ 7 } \) × 226 × 14 cm² = 9,944 cm²
Area of the hall uncovered = 9, 944 cm²

Question 4.
A school ground is in the shape of a circle with radius 103 m. Four tracks each of 3 m wide has to be constructed inside the ground for the purpose of track events. Find the cost of constructing the track at the rate of ₹ 50 per sq.m.

Solution:
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
= πR² – πr² sq.units
= π(R² – r²) sq.units
= \(\frac { 22 }{ 7 } \) [103² – 91²]
= \(\frac { 22 }{ 7 } \) [103 + 91] [103 – 91]m²
= \(\frac { 22 }{ 7 } \) × 194 × 12 = \(\frac { 51216 }{ 7 } \) = 7316.57 m²
∴ Area of 4 tracks = 7316.57 m²
Cost of constructing 7316.57 m² = ₹ 50
∴ Cost of constructing 7316.57 m² = ₹ 50 × 7316.57 = ₹ 3,65,828,57
Cost of constructing the track ₹ 3,65,828,57

Question 5.
The figure shown is the aerial view of the pathway. Find the area of the pathway.

Solution:
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 m² = 4000 m²
Area of the inner rectangle = l × b sq. unit = 70 × 40 m² = 2800 m²
Area of the pathway = Area of the outer rectangle
– Area of the inner rectangle
= 4000 – 2800 m² = 1200 m²
Area of the pathway = 1200 m²

Question 6.
A rectangular garden has dimensions 11 m × 8 m. A path of 2 m wide has to be constructed along its sides. Find the area of the path.
Solution:
Area of the rectangular garden L × B = 11 m × 8 m = 88 m²
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 m² = 28 m²
Area of the path = Area of the outer rectangular garden
– Area of the inner rectangle
= 88 m² – 28m² = 60 m²
Area of the path = 60 m²

Question 7.
A picture is painted on a ceiling of a marriage hall whose length and breadth are 18 m and 7 m respectively. There is a border of 10 cm along each of its sides. Find the area of the border.
Solution:
Length of the ceiling L = 18 m
Breadth of the ceiling B = 7 m
Area of the ceiling = L × B sq. units = 18 × 7 m² = 126 m²
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 m² = 121.04 m²
∴ Area of the border = Area of the ceiling
– Area of the ceiling without border
= 126 – 121.04 m² = 4.96 m²
Area of the border = 4.96 m²

Question 8.
A canal of width 1 m is constructed all along inside the field which is 24 m long and 15 m wide. Find (i) the area of the canal (ii) the cost of constructing the canal at the rate of ₹ 12 per sq.m.
Solution:
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 m² = 360 m²
(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 m² = 286 m²
Area of the canal = 360 – 286 = 74 m²
Cost of constructing 1 m² canal = ₹ 12
Cost of the constructing 74 m² canal = ₹ 12 × 74 = ₹ 888

Objective Type Question

Question 9.
The formula to find the area of the circular path is
(i) π(R² – r²) sq. units
(ii) πr² sq. units
(iii) 2πr² sq. units
(iv) πr² + 2r sq. units
Answer:
(i) π(R² – r2) sq. units

Question 10.
The formula used to find the area of the rectangular path is
(i) p(R² – r²) sq. units
(ii) (L × B) – (l × b) sq. units
(iii) LB sq. units
(iv) lb sq. units
Answer:
(ii) (L × B) – (l × b) sq. units

Question 11.
The formula to find the width of the circular path is
(i) (L – l) units
(ii) (B – b) units
(iii) (R – r) units
(iv) (r – R) units
Answer:
(iii) (R – r) units