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  ?
 A: The area of a sector of a circle with angle $\theta$ is  $$ \frac{\theta}{360}\pi r^2$$
For your smaller circle, the shaded area is \begin{align}\frac{360-\theta}{360}\pi r_1^2&= \frac{360-40}{360}\pi 7^2\\
&= \frac{320}{360}\times 49\pi\\
&= \frac 89 \times 49\pi\end{align}
For the larger circle we want to calculate the whole sector area and subtract the smaller white sector: 
\begin{align}\frac {\theta}{360}\pi r_2^2 - \frac{\theta}{360}\pi r_1^2 &= \frac{40}{360}\pi\times 14^2 - \frac{40}{360}\pi \times 7^2\\
&= \frac19\times 196\pi - \frac 19 \times 49\pi\\
&= \frac \pi 9 (196-49)\\
&= \frac \pi 9 \times 147\end{align}
Therefore the total shaded area is \begin{align}\frac 89 \times 49\pi+\frac \pi 9 \times 147&= \frac \pi 9(8\times 49+147)\\
&=\frac \pi 9 \times 539\\
&= \frac {539}9\pi\end{align}
A: 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.
A: Area of a circular sector: $$\frac{\pi r^2 \cdot \alpha}{360º}$$
where $\alpha =$ angle of the circular sector
With this, the shaded region of the smaller circle is $$\frac{\pi 7^2 \cdot 320º}{360º}$$
For the big circle, we have to substract the white area of the smaller circle from the circular sector of the big circle:
$$\frac{\pi 14^2 \cdot 40º}{360º}- \frac{\pi 7^2 \cdot 40º}{360º}$$ 
A: The way you've done it is not correct. We need to treat both the shaded areas as sectors of a circle (in the case of the outer one, a sector of the large circle minus a sector of a small circle)
The correct way as far as I know would be:
$\frac{40}{360}\pi(r_2^2 - r_1^2) + \frac{360-40}{360}\pi r_1^2$
You would simplify to answer.
