 Area of Circle NCERT Exercise 12.2 solution part one Class Ten Mathematics

# Area of Circle

## 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 360o in 60 minutes.

Hence, angle made in 5 minutes = 30o

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:

1. the length of the arc
2. area of the sector formed by the arc
3. 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 60o, 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