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Question

# A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of the circle. Find the area of major and minor segments of the circle.

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Solution

## Radius of the circle, r = 10 cm Area of sector OPRQ $=\frac{60°}{360°}×\mathrm{\pi }{r}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}×3.14×{\left(10\right)}^{2}\phantom{\rule{0ex}{0ex}}=52.33{\mathrm{cm}}^{2}$ 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 $=\frac{\sqrt{3}}{4}×{\left(\mathrm{Side}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}×{\left(10\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{100\sqrt{3}}{4}{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}=43.30{\mathrm{cm}}^{2}$ 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 $=\mathrm{\pi }{\left(10\right)}^{2}-9.03\phantom{\rule{0ex}{0ex}}=314-9.03\phantom{\rule{0ex}{0ex}}=304.97{\mathrm{cm}}^{2}$

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