Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4

Students can Download Maths Chapter 2 Measurements Ex 2.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4

Miscellaneous Practice Problems

Question 1.
A wheel of a car covers a distance of 3520 cm in 20 rotations. Find the radius of the wheel?
Solutions:
Distance covered by circular wheel in 20 rotation = 3520 cm
∴ Distance covered ini rotation = \(\frac { 3520 }{ 20 } \) cm = 176 cm
∴ Circumference of the wheel = 176 cm
∴ 2πr = 176
2 × \(\frac { -2 }{ 6 } \) × r = 176
r = \(\frac{176 \times 7}{2 \times 22}\)
r = 28 cm
Radius of the wheel = 28 cm

Question 2.
The cost of fencing a circular race course at the rate of ₹ 8 per metre is ₹2112. Find the diameter of the race course.
Solution:
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

Question 3.
A path 2 m long and 1 m broad is constructed around a rectangular ground of dimensions 120 m and 90 m respectively. Find the area of the path.
Solution:
Length of the rectangular ground l = 120 m
Breadth b = 90 m
Length of the path W1 = 2m
Length of the path W2 = 1m
Length of the ground with path L = 1 + 2 (W2) = 120 + 2(1) m
= 120 + 2 = 122 m
Breadth of the ground with path B = l + 2(W1) units
= 90 + 2(2) m = 90 + 4 m = 94 m
∴ Area of the path = (L × B) – (1 × b) sq. units
= (122 × 94) – (122 × 94) m2 = 668 m2
∴ Area of the path = 668 m2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4

Question 4.
The cost of decorating the circumference of a circular lawn of a house at the rate of ₹55 per metre is ₹16940. What is the radius of the lawn?
Solution:
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

Question 5.
Four circles are drawn side by side in a line and enclosed by a rectangle as shown below.
If the radius of each of the circles is 3 cm, then calculate:
(i) The area of the rectangle.
(ii) The area of each circle.
(iii) The shaded area inside the rectangle.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 2.4 1
Solution:
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 cm2
Area of the rectangle = 144 cm2

(ii) Area of the circle = πr2 sq. units
= \(\frac { 22 }{ 7 } \) × 3 × 3 cm2
= \(\frac { 198 }{ 7 } \) cm2
= 28.28 cm2

(iii) Area of the shaded area = Area of the rectangle – Area of the 4 circles
= 144 – (4 × \(\frac { 198 }{ 7 } \)) cm2 = 144 – \(\frac { 792 }{ 7 } \) cm2
= 144 – 113.14 cm2 = 30.85 cm2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4

Challenge Problems

Question 6.
A circular path has to be constructed around a circular lawn. If the outer and inner circumferences of the path are 88 cm and 44 cm respectively, find the width and area of the path.
Solution:
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 = πR2 – πr2 sq. units
= π (R + r) (R – r) = 66 × 7
Area of the path = 462cm2

Question 7.
A cow is tethered with a rope of length 35 m at the centre of the rectangular field of length 76 m and breadth 60 m. Find the area of the land that the cow cannot graze?
Solution:
Length of the field l = 76 m
Breadth of the field b = 60m
Area of the field A = l × b sq. units = 76 × 60 m2
Area of the field A = 4560 m2
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 = πr2 sq.units
π × 35 × 35 m2 = \(\frac { 22 }{ 7 } \) × 35 × 35 m2
= 3850 m2
Area of the land the cow cannot graze = Area of the field – Area that the cow can graze
= 4560 – 3860 m2 = 710 m2
Area of the land that the cow cannot graze = 710 m2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4

Question 8.
A path 5 m wide runs along the inside of the rectangular field. The length of the rectangular field is three times the breadth.of the field. If the area of the path is 500 m2 then find the length and breadth of the field.
Solution:
Let the length of the rectangular field = ‘L’ m
Breadth of the rectangular field = = ‘B’ m
Area of the rectangular field = (L × B) m2
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)
= 3B2 – 10B – 30B + 100
Area of the path = Area of outer rectangle
– Area of inner rectangle
= (L × B) – (3B2 – 10B – 30B + 100)
3B × B – (3B2 – 40B + 100)
= 3B2 – 3B2 + 40B – 100
Area of the path = 40B – 100
Given area of the path = 500 m2
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

Question 9.
A circular path has to be constructed around a circular ground. 1f the areas of the outer and inner circles are 1386 m2 and 616 m2 respectively, find the width and area of the path.
Solution:
Area of the outer circle = 1386 m2
πR2 = 1386m2
Area of the inner circle = 616 m2
πr2 = 616m2
Area of the path = Area of outer circle – Area of the inner circle
1386 m2 – 616 m2
Area of the path = 770m2
Also πR2 = 1386
R2 = \(\frac{1386 \times 7}{22}\)
R2 = 63 × 7
R2 = 9 × 7 × 7
R2 = 32 × 72
R = 3 × 7
Outer Radius R = 21 m
Again πr2 = 616
\(\frac { 22 }{ 7 } \) × r2 = 616
r2 = 28 × 7
r2 = 4 × 7 × 7
r2 = 22 × 72
r = 2 × 7
Inner radius r = 14m
Width of the path = Outer radius – Inner radius = 21 – 14
Width of the path = 7m

Question 10.
A goat is tethered with a rope of length 45 m at the centre of the circular grass land whose radius is 52 m. Find the area of the grass land that the goat cannot graze.
Solution:
Length of the rope = 45 m = Radius of the inner circle
∴ Area of the circular area that the goat graze = πr2 sq. units
= \(\frac { 22 }{ 7 } \) × 45 × 45 m2 = 6364.28 m2
Radius of the gross land = 52 m
Area of the grass land = \(\frac { 22 }{ 7 } \) × 52 × 52 = 8,498.28 m2
Area that the goat cannot graze
= Area of the outer circle – Area of the inner circle
= 8498.28 – 6364.28 = 2134 m2
Area of the goat cannot grass = 2134 m2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4

Question 11.
A strip of 4 cm wide is cut and removed from all the sides of the rectangular cardboard with dimensions 30 cm × 20 cm. Find the area of the removed portion and area of the remaining cardboard.
Solution:
Area of the outer rectangular cardboard
= L × B sq.units = 30 × 20 cm2 = 600 cm2
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 cm2 = 264 cm2
Area of the remaining cardboard = 264 cm2
Area of the removed portion = Area of outer rectangle
– Area of the inner rectangle
= 600 – 264 cm2
Area of the removed portion = 336 cm2

Question 12.
A rectangular field is of dimension 20 m × 15 m. Two paths run parallel to the sides of the rectangle through the centre of the field. The width of the longer path is 2m and that of the shorter path is 1 m. Find (i) the area of the paths (ii) the area of the remaining portion of the field (iii) the cost of constructing the roads at the rate of ₹ 10 per sq.m.
Solution:
Length of the rectangular field L = 20 m
Breadth B = 15m
Area = L × B
20 × 15 m2
Area of outer rectangle = 300 m2
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 2.4 2
Area of inner small rectangle = \(\frac { 19 }{ 2 } \) × \(\frac { 13 }{ 2 } \) = 61.75 cm2

(i) Area of the path = Area of the outer rectangle
– Area of 4 inner small rectangles
= 300 – 4(61.75) = 300 – 247 = 53 m2
Area of the paths = 53 m2

(ii) Area of the remaining portion of the field
= Area of the outer rectangle – Area of the paths
= 300 – 53 m2 = 247 m2
Area of the remaining portion = 247 m2

(iii) Cost of constructing 1 m2 road = ₹10
∴ Cost of constructing 53 m2 road = ₹10 × 53 = ₹530
∴ Cost of constructing road = ₹530

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