Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Students can Download Maths Chapter 2 Measurements Ex 2.2 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 2.2

Question 1.
Find the area of the dining table whose diameter is 105 cm.
Solution:
Diameter of the dinig table (d) = 105 cm
∴ Radius r = \(\frac { d }{ 2 } \) = \(\frac { 105 }{ 2 } \) cm
Area of the circle = π r2 = \(\frac { 22 }{ 7 } \) × \(\frac { 105 }{ 2 } \) × \(\frac { 105 }{ 2 } \) = 8662.5 sq.cm
Area of the dinning table = 8662.5 cm2

Question 2.
Calculate the area of the shotput ring whose diameter is 2.135 m.
Solution:
Radius of the shotput ring r = \(\frac { d }{ 2 } \) = \(\frac { 2.135 }{ 2 } \) m
Area of the circle = π r2
= \(\frac { 22 }{ 7 } \) × \(\frac { 2.135 }{ 2 } \) × \(\frac { 2.135 }{ 2 } \)
= \(\frac { 25.07 }{ 7 } \) = 3.581 m2
∴ Area of the shotput ring = 3.581 m2

Question 3.
A sprinkler placed at the centre of a flower garden sprays water covering a circular area. If the area watered is 1386 cm2, find its radius and diameter.
Solution:
Area of the Circle = π r2 sq.units
Area of the circular portion watered = 1386 cm2
π r2 = 1386
\(\frac { 22 }{ 7 } \) × r2 = 1386
r2 = 1386 × \(\frac { 7 }{ 22 } \) = 63 × 7 = 9 × 7 × 7
r2 = 32 × 72
r = 3 × 7
Radius (r) = 21 cm
Diameter (d) = 2 r = 2 × 21 cm
Diameter (d) = 42 cm

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Question 4.
The circumference of a circular park is 352 m. Find the area of the park.
Solution:
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 r2 = \(\frac { 22 }{ 7 } \) × 56 × 56 sq.units
= 22 × 8 × 56 = 9856 m2
∴ Area of the Circular park = 9856 m2

Question 5.
In a grass land, a sheep is tethered by a rope of length 4.9 m. Find the maximum area that the sheep can graze.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 2.2 1
Solution:
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 r2 sq.units
= \(\frac { 22 }{ 7 } \) × 4.9 × 4.9 = 22 × 0.7 × 4.9 = 75.46
∴ Area that the sheep can graze = 75.46 m2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Question 6.
Find the length of the rope by which a bull must be tethered in order that it may be able to graze an area of 2464 m2.
Solution:
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 m2
π r2 = 2464 m2
\(\frac { 22 }{ 7 } \) × r2 = 2464
r2 = 2464 × \(\frac { 7 }{ 22 } \) = 122 × 7 = 16 × 7 × 7
r2 = 42 × 72
r = 4 × 7 = 28 m
length of the rope r = 28 m

Question 7.
Lalitha wants to buy a round carpet of radius is 63 cm for her hall. Find the area that will be covered by the carpet.
Solution:
Radius of the round carpet = 63 cm
Area covered by the round carpet = πr2 sq units
A = \(\frac { 22 }{ 7 } \) × 63 × 63 = 22 × 9 × 63 = 12474 cm2
Area covered by the round carpet = 12,474 cm2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Question 8.
Thenmozhi wants to level her circular flower garden whose diameter is 49 m at the rate of ₹150 per m2 Find the cost of levelling.
Solution:
Diamter of the circular garden d = 49 m
Radius r = \(\frac { d }{ 2 } \) = \(\frac { 49 }{ 2 } \) m
Area of the circular garden = πr2 sq units
= \(\frac { 22 }{ 7 } \) × \(\frac { 49 }{ 2 } \) × \(\frac { 49 }{ 2 } \) m2 = 1,886.5 m2
Cost of levelling a m2 area = ₹ 150
∴ Cost of levelling 1886.5 m2 = ₹ 150 × 1886.5 = ₹ 2,82,975
Cost of levelling the flower garden = ₹ 2,82,975

Question 9.
The floor of the circular swimming pool whose radius is 7 m has to be cemented at the rate of ₹ 18 per m2. Find the total cost of cementing the floor.
Solution:
Radius of the circular swimming pool r = 7 m
Area of the circular swimming pool A = πr2 sq. units
= \(\frac { 22 }{ 7 } \) × 7 × 7 m2 = 154 m2
Cost of cementing a m2 floor = ₹ 18.
Cost of cementing 154 m2 floor = ₹ 18 × 154 = ₹ 2,772

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Objective Type Questions

Question 10.
The formula used to find the area of the circle is
(i) 47πr2
(ii) πr2
(iii) 2πr2
(iv) πr2 + 2r
Answer:
(ii) πr2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Question 11.
The ratio of the area of a circle to the area of its semicircle is
(i) 2 : 1
(ii) 1 : 2
(iii) 4 : 1
(iv) 1 : 4
Answer:
(i) 2 : 1

Question 12.
Area of circle of radius ‘n’ units is
(i) 2πrp sq.units
(ii) πm2 sq.units
(iii) πr2 sq.units
(iv) πn2 sq.units
Answer:
(iv) πn2 sq.units

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