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°.
Answer: 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.
Answer: 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.
Answer: 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.
Answer: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
Answer: 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.
Answer: 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`