Students can Download Maths Chapter 2 Measurements Ex 2.1 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 2.1
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:
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
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
30 frames can be made.
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
Answer:
(i) 2πr units
Question 10.
In the formula, C = 2πr, ‘r’ refers to
(i) circumference
(ii) area
(iii) rotation
(iv) radius
Answer:
(iv) radius
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
Answer:
(i) 41cm
Question 12.
Circumference of a circle is always
(i) three times of its diameter
(ii) more than three times of its diameter
(iii) less than three times of its diameter
(iv) three times of its radius
Answer:
(ii) more than three times of its diameter