Find the area of the shaded region of two circle with the radius of $r_1$ and $r_2$ In the given figure , $O$  is the center of the circle and $r_1 =7cm$,$r_2=14cm,$ $\angle AOC =40^{\circ}$. Find the area  of the shaded region

My attempt:  Area  of shaded  region $=\pi r^2_2 - \pi r^2_1= \pi( 196-49)= 147\pi$
Is it true  ?

Use the formula for the area of a sector (the angle is measured in radians) and the formula for the area of a circle:
$$
A=\frac{1}{2}r^2\theta
$$
$$
A=\pi r^2
$$
$40^\circ$ in radians would be:
$$
40^\circ=\frac{40\pi}{180}
$$
The area of the top peice:
$$
A_1=\frac{1}{2}r_2^2\cdot \frac{40\pi}{180} - \frac{1}{2}r_1^2\cdot\frac{40\pi}{180}=\frac{\pi}{9}\left(r_2^2-r_1^2\right)
$$
The area of the bottom piece:
$$
A_2=\pi r_1^2-\frac{1}{2}r_1^2\cdot\frac{40\pi}{180}=\frac{8\pi}{9}r_1^2
$$
Thus, the area of the shaded region would be:
$$
A_1+A_2=\frac{\pi}{9}\left(r_2^2-r_1^2\right)+\frac{8\pi}{9}r_1^2=\frac{\pi}{9}\left(r_2^2+7r_1^2\right)=\frac{\pi}{9}\left(14^2+7\cdot7^2\right)=\frac{539\pi}{9}
$$
Answer: $A=\frac{539\pi}{9}$ square units.
