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.

Answer 1

Radius of the circle, r = 10 cm

Area of sector OPRQ
$$\begin{gathered} =\frac{60°}{360°}\times \mathrm{\pi }{r}^2 \\ =\frac{1}{6}\times 3.14\times {\left(10\right)}^2 \\ =52.33{\mathrm{cm}}^2 \end{gathered}$$

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
$$\begin{gathered} =\frac{\sqrt{3}}{4}\times {\left(\mathrm{Side}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times {\left(10\right)}^2 \\ =\frac{100\sqrt{3}}{4}{\mathrm{cm}}^2 \\ =43.30{\mathrm{cm}}^2 \end{gathered}$$

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
$$\begin{gathered} =\mathrm{\pi }{\left(10\right)}^2-9.03 \\ =314-9.03 \\ =304.97{\mathrm{cm}}^2 \end{gathered}$$