Master Class 10 Math Area Related To Circle Ex 12.2 NCERT

In this section, we provide detailed and step-by-step NCERT solutions for Class 10 Maths Chapter 12 – Areas Related to Circles, Exercise 12.2. This exercise mainly focuses on the application of the formulas for the area of a sector and the length of an arc of a circle. These concepts are essential for solving real-life problems involving circular paths, wheels, slices of pizza, and more.

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Class 10 Math Area Related to Circle Ex 12.2 – Textbook

Question 1.
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:

Radius (r) = 6 cm
Angle (θ) = 60°

Formula:
Area of sector = (θ / 360) × π × r²
= (60 / 360) × (22 / 7) × 6 × 6
= (1 / 6) × (22 / 7) × 36
= (22 × 36) / (6 × 7)
= 792 / 42
= 18.86 cm² (approximately)

Ex 12.2 Class 10 Maths Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Step 1: Find the radius

Using:
2 × π × r = 22
Take π = 22/7

2 × (22/7) × r = 22
(44/7) × r = 22
r = 22 × (7/44) = 7/2 = 3.5 cm

Step 2: Area of a full circle

Area = π × r² = (22/7) × (3.5)²
= (22/7) × 12.25
= (22 × 12.25) / 7
= 269.5 / 7 ≈ 38.5 cm²

Step 3: Area of a quadrant

A quadrant is 1/4 of the full circle:
= 38.5 ÷ 4 = 9.625 cm²

Ex 12.2 Class 10 Maths Question 3.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
The length of the minute hand of a clock is 14 cm.
We are asked to find the area swept by the minute hand in 5 minutes.

In 60 minutes, the minute hand completes a full circle, which is 360 degrees.
So, in 1 minute, it sweeps 6 degrees.
In 5 minutes, it sweeps:
6 × 5 = 30 degrees

Now, we use the formula for the area of a sector:

Area = (θ / 360) × π × r²
Here,
θ = 30 degrees
r = 14 cm
π ≈ 3.14

Area = (30 / 360) × 3.14 × (14)²
= (1 / 12) × 3.14 × 196
= (1 / 12) × 615.44
= 51.29 cm²

Ex 12.2 Class 10 Maths Question 4.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
(i) minor segment
(ii) major segment (Use π = 3.14)
Solution:
Given:
Radius of the circle = 10 cm
Angle subtended at the centre = 90°
π = 3.14

(i) Area of the minor segment

First, find the area of the sector:

Area of sector = (θ / 360) × π × r²
= (90 / 360) × 3.14 × 10²
= (1 / 4) × 3.14 × 100
= 78.5 cm²

Now, find the area of the triangle formed by the two radii and the chord. Since the angle is 90°, it forms a right-angled isosceles triangle.

Area of triangle = (1 / 2) × 10 × 10 × sin(90°)
= (1 / 2) × 10 × 10 × 1
= 50 cm²

Area of minor segment = Area of sector – Area of triangle
= 78.5 – 50
= 28.5 cm²

(ii) Area of the major segment

Total area of the circle = π × r²
= 3.14 × 10²
= 3.14 × 100
= 314 cm²

Area of major segment = Total area of circle – Area of minor segment
= 314 – 28.5
= 285.5 cm²

Ex 12.2 Class 10 Maths Question 5.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) length of the arc.
(ii) area of the sector formed by the arc.
(iii) area of the segment formed by the corresponding chord.
Solution:
Given:
Radius of the circle = 21 cm
Angle subtended at the centre = 60°
π = 22/7

(i) Length of the arc:
Length of arc = (θ / 360) × 2 × π × r
= (60 / 360) × 2 × (22 / 7) × 21
= (1 / 6) × 2 × (22 / 7) × 21
= (1 / 6) × (44 × 3)
= (1 / 6) × 132
= 22 cm

(ii) Area of the sector:
Area of sector = (θ / 360) × π × r²
= (60 / 360) × (22 / 7) × 21 × 21
= (1 / 6) × (22 / 7) × 441
= (22 × 441) / (6 × 7)
= 9702 / 42
= 231 cm²

(iii) Area of the segment:
Area of triangle = (1 / 2) × r² × sin(θ)
= (1 / 2) × 21 × 21 × sin(60°)
= (1 / 2) × 441 × 0.866
= 220.5 × 0.866
= 190.95 cm²

Area of segment = Area of sector – Area of triangle
= 231 – 190.95
= 40.05 cm²

Ex 12.2 Class 10 Maths Question 6.
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)
Solution:
Given:
Radius of the circle = 15 cm
Angle subtended at the centre = 60°
π = 3.14
√3 = 1.73

(i) Area of the sector:
Area of sector = (60 / 360) × 3.14 × 15 × 15
= (1 / 6) × 3.14 × 225
= (1 / 6) × 706.5
= 117.75 cm²

(ii) Area of the triangle:
Area of triangle = (√3 / 4) × side²
= (1.73 / 4) × 15 × 15
= 0.4325 × 225
= 97.31 cm²

(iii) Area of the minor segment:
Area of minor segment = Area of sector – Area of triangle
= 117.75 – 97.31
= 20.44 cm²

(iv) Area of the major segment:
Total area of the circle = 3.14 × 15 × 15 = 706.5 cm²
Area of major segment = 706.5 – 20.44 = 686.06 cm²

Ex 12.2 Class 10 Maths Question 7.
A chord of a circle of the radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73).
Solution:
Given:
Radius of the circle = 12 cm
Angle subtended at the centre = 120°
π = 3.14
√3 = 1.73

Step 1: Area of the sector
Area of sector = (120 / 360) × 3.14 × 12 × 12
= (1 / 3) × 3.14 × 144
= 1.0467 × 144
= 150.73 cm²

Step 2: Area of the triangle
Area of triangle = (1 / 2) × 12 × 12 × sin(120°)
= (1 / 2) × 144 × 0.865
= 72 × 0.865
= 62.36 cm²

Step 3: Area of the segment
Area of segment = Area of sector – Area of triangle
= 150.73 – 62.36
= 88.37 cm²

Ex 12.2 Class 10 Maths Question 8.
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find:
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2 Q8
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)

Solution:
Given:
Side of the square field = 15 m
Length of rope = 5 m
π = 3.14

(i) Area the horse can graze:
Since the horse is tied at one corner, it grazes in a quarter of a circle.

Area = (1/4) × π × r²
= (1/4) × 3.14 × 5 × 5
= (1/4) × 3.14 × 25
= (1/4) × 78.5
= 19.625 m²

So, grazing area = 19.63 m²

(ii) If rope length is 10 m:
Area = (1/4) × 3.14 × 10 × 10
= (1/4) × 3.14 × 100
= (1/4) × 314
= 78.5 m²

Increase in grazing area = 78.5 – 19.63 = 58.87 m²

Ex 12.2 Class 10 Maths Question 9.
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure.
Find:

(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Solution:
Given:
Diameter of the circular brooch = 35 mm
Radius = 35 / 2 = 17.5 mm
The brooch has 5 diameters, dividing the circle into 10 equal sectors
π = 3.14

(i) Total length of silver wire required:
The wire is used to make the circular boundary and 5 diameters.

Length of circular boundary = π × diameter
= 3.14 × 35
= 109.9 mm

Length of 5 diameters = 5 × 35 = 175 mm

Total length of wire = 109.9 + 175 = 284.9 mm

(ii) Area of each sector:
Area of the circle = π × r²
= 3.14 × 17.5 × 17.5
= 3.14 × 306.25
= 961.63 mm²

Since the circle is divided into 10 equal sectors:
Area of each sector = 961.63 / 10 = 96.16 mm²

Ex 12.2 Class 10 Maths Question 10.
An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2 Q10
Solution:
Given:

The umbrella has 8 ribs (equally spaced), forming 8 equal sectors.
Radius of the umbrella = 45 cm
π = 3.14

We are to find the area between two consecutive ribs, which is the area of one sector out of 8.

Step 1: Find angle of one sector
Since the umbrella is a full circle (360°) and has 8 ribs:

Angle of one sector = 360° / 8 = 45°

Step 2: Use the formula for the area of a sector
Area of sector = (θ / 360) × π × r²
= (45 / 360) × 3.14 × 45 × 45
= (1 / 8) × 3.14 × 2025
= 0.125 × 3.14 × 2025
= 0.125 × 6358.5
= 794.81 cm²

Ex 12.2 Class 10 Maths Question 11.
A car has two wipers which do not overlap.
Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution:
Given:

Length of each wiper blade = 25 cm
Angle swept by each blade = 115°
The wipers do not overlap
There are 2 wipers
π = 3.14

Step 1: Area swept by one wiper

Area of sector = (θ / 360) × π × r²
= (115 / 360) × 3.14 × 25 × 25
= (115 / 360) × 3.14 × 625
= 0.3194 × 3.14 × 625
= 0.3194 × 1962.5
= 626.15 cm²

Step 2: Total area cleaned by two wipers

= 2 × 626.15
= 1252.30 cm²

Ex 12.2 Class 10 Maths Question 12.
To warn ships for underwater rocks, a lighthouse spreads a red colored light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Solution:
Given:

Angle of the sector = 80°
Radius (distance light reaches) = 16.5 km
π = 3.14

We are to find the area of the sector (sea region warned by the lighthouse).

Step 1: Use the formula for the area of a sector
Area of sector = (θ / 360) × π × r²
= (80 / 360) × 3.14 × 16.5 × 16.5
= (2 / 9) × 3.14 × 272.25
= 0.2222 × 3.14 × 272.25
= 0.2222 × 854.87
= 189.97 km²

Ex 12.2 Class 10 Maths Question 13.
A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹0.35 per cm². (Use √3= 1.7)
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2 Q13
Solution:
Given:
Radius of the round table cover = 28 cm
Number of equal designs = 6
Each design is a segment of 60°
π = 3.14
√3 = 1.7
Cost of making design = ₹0.35 per cm²

Step 1: Area of the sector
Area of sector = (60 / 360) × 3.14 × 28 × 28
= (1 / 6) × 3.14 × 784
= 409.96 cm²

Step 2: Area of the triangle inside the sector
Area of triangle = (√3 / 4) × side²
= (1.7 / 4) × 28 × 28
= 0.425 × 784
= 333.2 cm²

Step 3: Area of the segment
Area of segment = Area of sector – Area of triangle
= 409.96 – 333.2
= 76.76 cm²

Step 4: Area of 6 such designs
= 6 × 76.76
= 464.82 cm²

Step 5: Cost of making the designs
= 464.82 × 0.35
= ₹162.69

Ex 12.2 Class 10 Maths Question 14.
Tick the correct answer in the following: Area of a sector of angle p (in degrees) of a circle with radius R is

Solution:
Areas Related To Circles Class 10 Maths NCERT Solutions Ex 12.2 PDF Q14

The questions and solutions provided on this page are based on the NCERT Class 10 Mathematics textbook – Chapter 12: Areas Related to Circles. For detailed study and official content, you can refer to the NCERT textbook available on the official NCERT website.