# Class 10 NCERT Solutions- Chapter 12 Areas Related to Circles – Exercise 12.2 | Set 1

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

**Solution:**

Given: r=66cm and ∅=60°

_{Area of sector=∅/(360°) *πr}2=60/360*22/7*6*6

=132/7

**Question 2. Find the area of a quadrant of a circle whose circumference is 22 cm. Quadrant of circle is sector making 90°.**

**Solution:**

Given: circumference of circle=22cm and ∅=90°

To find r=?

2πr=22

Radius =r=22/2πr cm=7/2cm

_{Area of quadrant =∅/(360°) *πr}2=90/360*22/7*7/2*7/2

=77/8cm2

**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:**

Let assume, Minute hand of clock acts as radius of the circle.

Angle rotated by min (hand in 5minutes)=∅=360/60*5=30°

Radius=r=14cm

Area of swept middle hand=

_{=∅/(360°) *πr}2=30/360*22/7*14*14

_{=154/3cm}2

_{Area swept by the minute hand in 5min=154/3cm}2

**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 sector. (Use **π** = 3.14)**

**Solution:**

Radius =r=10cmMajor segment is =360°-90°=270°

(i)Area of minor segment =Area of sector-Area of triangle=∅/(360°) *πr2-1/2*h*b

=90/306*3.14*10*10-1/2*10*10

=314/4-50

=78.5-50

_{Area of minor segment =28.5cm}2

_{(ii)}_{ Area of major sector=∅/(360°) *πr}2=270/360*3.14*10*10

_{=3*314/4=235.5cm}2

_{Area of major segment=235.5cm}2

**Question 5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:**

**(i) the length of the arc**

**(ii) area of the sector formed by the arc**

**(iii) area of the segment formed by the corresponding chord**

**Solution:**

(i)Radius=r=21cmLength of arc=∅/(360°) *2πr

60/360*2*22/7*21

=22cm

_{(ii)}_{ Area of sector=∅/(360°) *πr}260/360*22/7*21*21

_{11*21=231cm}2

(iii)Area of segment =Area of sector -Area of triangle

_{=231-√3/4(side)}2=231-1.73/4*21*21

=231-762.93/4

=231-190.73

_{=40.27cm}2

**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:**

Radius of circle=15cm

∆AOB is isosceles

∴∠A = ∠B

=∠A+∠B+C=180°

=2∠A=180°-60°

=∠A=120°/2

=∠A=60°

Area of minor segment =Area of sector-Area of triangle

_{=∅/(360°) *πr}2-_{√3/4(side)}2=(60°)/360*3.14*15*15-1.73/4

=706.5/6-389.25/4

=117.75-97.31

=

_{20.44cm}2Area of major segment-Area of circle-Area of minor segment

_{=πr}2-20.44=3.14*15*15-20.44

_{=686.06cm}2

**Question 7. A chord of a circle of 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:**

Radius=r=12cm

Area of triangle=1/2*base*height

Area of segment=Area of sector -area of triangle

=∅/(360°)*π*r

^{2}-1/2(side)^{2}*sin∅=120°/360°*3.14-1/2(side)

^{2}*sin120°∅=150.72-6*12*sin60°

=150.72-6*12*√3/2

=150.72-36*1.73

=150.72-62.28

_{=88.44cm}2