# Area of Circle

## NCERT Exercise

### 12.2 Part 1

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

**Solution:** Area of sector

`=(θ)/(360°)πr^2`

`=(60°)/(360°)xx(22)/(7)xx6^2`

`=18.84 sq cm`

Question 2: Find the area of a quadrant of a circle whose circumference is 22 cm.

**Solution:** Radius can be calculated as follows:

`r=text(circumference)/(2π)`

`=(22xx7)/(2xx22)=3.5 cm`

Area of quadrant can be calculated as follows:

`=(1)/(4)xx(22)/(7)xx3.5^2`

`=9.625 sq cm`

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 minute hand makes an angle of 360^{o} in 60 minutes.

Hence, angle made in 5 minutes = 30^{o}

Area of sector

`=(θ)/(360°)πr^2`

`=(30°)/(360°)xx(22)/(7)xx14^2`

`=51.33 sq cm`

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.

**Solution:**Area of minor sector

`=(1)/(4)xx πr^2`

`=(1)/(4)xx πxx10^2`

`=25 π sq cm`

Area of major sector

`=(3)/(4)xx πr^2`

`=(3)/(4)xx πxx10^2`

`=75 π sq cm`

Area of triangle formed by two radii

`=(1)/(2)xxtext(base)xxtext(height)`

`=(1)/(2)xx10xx10=50 sq cm`

Area of minor segment = Area of minor sector – area of triangle

`= 25π - 50 = 25(π - 2)`

`= 25(3.14 – 2) = 25 xx 1.14 = 28.5 sq cm`

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

- the length of the arc
- area of the sector formed by the arc
- area of the segment formed by the corresponding chord

**Solution:** Length of Arc can be calculated as follows:

`=(θ)/(360°)xx2πr`

`=(60°)/(360°)xx2xx(22)/(7)xx21`

`=22 cm`

Area of corresponding sector can be calculated as follows:

`=(60°)/(360°)xxπr^2`

`=(1)/(6)xx(22)/(7)xx21^2`

`=231 sq cm`

As the angle made by radii is 60^{o}, so this is an equilateral triangle; because angles opposite to both radii will be equal.

Area of equilateral triangle can be calculated as follows:

`sqrt3/4xxtext(side)^2`

`=sqrt3/4xx21^2`

`=190.953 sq cm`

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.

**Solution:** Let us use previous figure to find area of triangle

`=sqrt3/4xxtext(side)^2`

`=sqrt3/4xx15^2`

`=97.425 sq cm`

Area of circle `= πr^2 = π15^2 = 225π = 706.5 sq cm`

Area of minor sector

`=(1)/(6)xx225π`

`=117.75 sq cm`

Area of minor segment `= 117.75 – 97.425 =20.325 sq cm`

Area of major segment `= 706.5 – 20.325 = 686.175 sq cm`