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प्रश्न
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 `sqrt3 = 1.73`]
उत्तर
Radius (r) of circle = 15 cm
Area of sector OPRQ = `60^@/360^@xxpir^2`
= `1/6 xx 3.14 xx (15)^2`
= 117.75 cm2
In ΔOPQ,
∠OPQ = ∠OQP (As OP = OQ)
∠OPQ + ∠OQP + ∠POQ = 180°
2 ∠OPQ = 120°
∠OPQ = 60°
ΔOPQ is an equilateral triangle.
Area of ΔOPQ = `(sqrt3)/4 xx (r)^2`
`= sqrt3/4 xx (15)^2`
= `(225sqrt3)/4 "cm"^2`
`= 56.25sqrt3`
97.3125 cm2
Area of segment PRQ = Area of sector OPRQ − Area of ΔOPQ
= 117.75 − 97.3125
= 20.4375 cm2
Area of major segment PSQ = Area of circle − Area of segment PRQ
= πr2 − 20.4375
= π × (15)2 − 20.4375
= 3.14 × 225 − 20.4375
= 706.5 − 20.4375
= 686.0625 cm2
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