Question

In the given figure, O is the centre of the circle. m ( arc PQR) = 60^° OP = 10 cm. Find the area of the shaded region.

( $\mathrm{\pi }$= 3.14, $\sqrt{3}$= 1.73)

Answer 1

Radius of the circle, r = 10 cm

m(arc PQR) = ∠POR = θ = 60º

∴ Area of the shaded region = Area of segment PQR

$$\begin{gathered} =r^2\left(\frac{\mathrm{\pi }\theta }{360°}-\frac{\sin \theta }{2}\right) \\ ={\left(10\right)}^2\left(\frac{3.14\times 60°}{360°}-\frac{\sin 60°}{2}\right) \\ =100\times \left(\frac{3.14}{6}-\frac{\sqrt{3}}{4}\right) \\ =\frac{314}{6}-\frac{173}{4}\left(\sqrt{3}=1.73\right) \\ =52.33-43.25 \\ =9.08{\mathrm{cm}}^2 \end{gathered}$$
Thus, the area of the shaded region is 9.08 cm².