Question

In the given figure, two concentric circles with centre O have radii 21 cm and 42 cm. If ∠AOB = 60°, find the area of the shaded region. $\left[\mathrm{Use}\mathrm{\pi }=\frac{22}{7}\right]$ [CBSE 2017]

Answer 1

Given: Radius of the inner circle with radius OC, r = 21 cm
Radius of the inner circle with radius OA, R = 42 cm
∠AOB = 60°

Area of the circular ring

$$\begin{gathered} =\mathrm{\pi }{R}^2-\mathrm{\pi }{r}^2 \\ =\mathrm{\pi }\left[R^2-r^2\right] \\ =\mathrm{\pi }\left[{42}^2-{21}^2\right]{\mathrm{cm}}^2 \end{gathered}$$

Area of ACDB = area of sector AOB − area of COD
$$\begin{gathered} =\frac{60}{360}\times \mathrm{\pi }\times R^2-\frac{60}{360}\times \mathrm{\pi }\times r^2 \\ =\frac{60}{360}\times \mathrm{\pi }\left[R^2-r^2\right] \\ =\frac{60}{360}\times \mathrm{\pi }\left[{42}^2-{21}^2\right] \end{gathered}$$

Area of shaded region = area of circular ring − area of ACDB
$$\begin{gathered} =\mathrm{\pi }\left[{42}^2-{21}^2\right]-\frac{60}{360}\mathrm{\pi }\left[{42}^2-{21}^2\right] \\ =\mathrm{\pi }\left[{42}^2-{21}^2\right]\left[1-\frac{60}{360}\right] \\ =\frac{22}{7}\left(42-21\right)\left(42+21\right)\times \frac{300}{360} \\ =3465{\mathrm{cm}}^2 \end{gathered}$$