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प्रश्न
A chord PQ of a circle of radius 10 cm substends an angle of 60° at the centre of circle. Find the area of major and minor segments of the circle.
उत्तर
Radius of the circle, r = 10 cm
Area of sector OPRQ
`= 60^@/360^@ xx pir^2`
`= 1/6 xx 3.14 xx (10)^2`
`= 52.33 cm^3`
In ΔOPQ,
∠OPQ = ∠OQP (As OP = OQ)
∠OPQ + ∠OQP + ∠POQ = 180°
2∠OPQ = 120°
∠OPQ = 60°
ΔOPQ is an equilateral triangle.
So, area of ΔOPQ
`= sqrt3/4 xx ("Side")^2`
`= sqrt3/4 xx (10)^2`
`=( 100sqrt3)/4 cm^2`
=43.30 cm2
Area of minor segment PRQ
= Area of sector OPRQ − Area of ΔOPQ
= 52.33 − 43.30
= 9.03 cm2
Area of major segment PSQ
= Area of circle − Area of minor segment PRQ
=π(10)2 − 9.03
=314 − 9.03
= 304.97 cm2
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