Find the area of the shaded region in fig., if radii of the two concentric circles with centre $\mathrm{O}$ are $7 \mathrm{~cm}$ and $14 \mathrm{~cm}$ respectively and $\angle \mathrm{AOC}=40^{\circ}$.
Radius of the outer circle $=14 \mathrm{~cm}$ and $\theta=40^{\circ}$
$\therefore \quad$ Area of the sector $\mathrm{AOC}$
$=\frac{40^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14 \mathrm{~cm}^{2}$
$=\frac{1}{9} \times 22 \times 2 \times 14 \mathrm{~cm}^{2}=\frac{616}{9} \mathrm{~cm}^{2}$
Radius of the inner circle $=7 \mathrm{~cm}$ and $\theta=40^{\circ}$
$\therefore$ Area of the sector BOD
$=\frac{40^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 7 \times 7 \mathrm{~cm}^{2}$
$=\frac{1}{9} \times 22 \times 7 \mathrm{~cm}^{2}=\frac{154}{9} \mathrm{~cm}^{2}$
Now, area of the shaded region
= Area of sector AOC – Area of sector BOD
$=\frac{\mathbf{6 1 6}}{\mathbf{9}}-\frac{\mathbf{1 5 4}}{\mathbf{9}} \mathrm{cm}^{2}=\frac{\mathbf{1}}{\mathbf{9}}(616-154) \mathrm{cm}^{2}$
$=\frac{1}{9} \times 462 \mathrm{~cm}^{2}=\frac{154}{3} \mathrm{~cm}^{2}$