# Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.1

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## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.1

Question 1.
Find the missing values in the following table for the circles with radius (r), diameter (d) and Circumference (C). Solution:
(i) Given radius r = 15cm
∴ diameter d = 2 × 15 = 30 cm
Circumference C = π d units
= $$\frac { 22 }{ 7 }$$ × 30 = $$\frac { 660 }{ 7 }$$ = 94.28 cm

(ii) Given circumference C = 1760 cm
2πr = 1760
2 × $$\frac { 22 }{ 7 }$$ × r = 1760
r = $$\frac{1760 \times 7}{2 \times 22}$$ = $$\frac{160 \times 7}{2 \times 2}$$ = 40 × 7 = 280 cm
diameter = 2 × r
= 2 × 280 = 560 cm

(iii) diameter d = 24m
radius r = $$\frac { d }{ 2 }$$ = $$\frac { 24 }{ 2 }$$ = 12 m
Circumference C = 2 π r units
= 2 × $$\frac { 22 }{ 7 }$$ × 12 = $$\frac { 528 }{ 7 }$$ = 75.4 m

Tabulating the results  Question 2.
Diameters of different circles are given below. Find their circumference (Take π = $$\frac { 22 }{ 7 }$$ )
(i) d = 70cm
(ii) d = 56m
(iii) d = 28mm
Solution:
(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

Question 3.
Find the circumference of the circles whose radii are given below.
(i) 49 cm
(ii) 91 mm
Solution:
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 Question 4.
The diameter of a circular well is 4.2 m. What is its circumference?
Solution:
Given the diameter d = 4.2 m
Circumference C = π d units = $$\frac { 22 }{ 7 }$$ × 4.2 m = 22 × 0.6 = 13.2 m

Question 5.
The diameter of the bullock cart wheel is 1.4 m. Find the distance covered by it in 150 rotations?
Solution:
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

Question 6.
A ground is in the form of a circle whose diameter is 350 m. An athlete makes 4 revolutions. Find the distance covered by the athlete.
Solution:
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 Question 7.
A wire of length 1320 cm is made into circular frames of radius 7 cm each. How many frames can be made?
Solution:
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 Question 8.
A Rose garden is in the form of circle of radius 63 m. The gardener wants to fence it at the rate of ₹ 150 per metre. Find the cost of fencing?
Solution:
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 Objective Type Questions

Question 9.
Formula used to find the circumference of a circle is
(i) 2πr units
(ii) πr2 + 2r units
(iii) πr2 sq. units
(iv) πr3 cu. units
(i) 2πr units

Question 10.
In the formula, C = 2πr, ‘r’ refers to
(i) circumference
(ii) area
(iii) rotation

Question 11.
If the circumference of a circle is 82π, then the value of ‘r’ is
(i) 41cm
(ii) 82 cm
(iii) 21cm
(iv) 20 cm 