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.
Radius of the circle, r = 10 cm
Area of sector $O P R Q$
$=\frac{60^{\circ}}{360^{\circ}} \times \pi r^{2}$
$=\frac{1}{6} \times 3.14 \times(10)^{2}$
$=52.33 \mathrm{~cm}^{2}$
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
$=\frac{\sqrt{3}}{4} \times(\text { Side })^{2}$
$=\frac{\sqrt{3}}{4} \times(10)^{2}$
$=\frac{100 \sqrt{3}}{4} \mathrm{~cm}^{2}$
$=43.30 \mathrm{~cm}^{2}$
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
$=\pi(10)^{2}-9.03$
$=314-9.03$
$=304.97 \mathrm{~cm}^{2}$
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