Question

In the figure, find the area of the shaded region, enclosed between two concentric circles of radii $7$ cm and $14$ cm where $\angle AOC={40}^{\circ }.$ Consider $\pi =\frac{22}{7}.$

Answer 1

Given: Radius of inner circle $=7$ cm

Radius of outer circle $=14$ cm

Angle made by sector = ${40}^o$

Area of sector OAC = $\frac{{40}^o}{{360}^o}\times \pi r^2$

$=\frac{1}{9}\times \times \frac{22}{7}\times {14}^2=68.44$

Area of sector OBD = $\frac{{40}^o}{{360}^o}\times \pi r^2$

$=\frac{1}{9}\times \times \frac{22}{7}\times 7^2=17.11$

Area of the small region $ABDC$
$=$ Area of the small sector $OAC$ $–$ Area of the small sector $OBD$ $=(68.44-17.11)c{m}^2$

Area of shaded region $ABDC=$ Area of outer circle $-$ Area of inner circle $-$ area of small region $ABCD$

Area of shaded region $ABDC=\pi \times {14}^2-\pi \times 7^2-(68.44-17.11)c{m}^2$

Area of shaded region $ABDC=410.484$ cm${}^2$