[Math] Find the area of the shaded region of two circle with the radius of $r_1$ and $r_2$

Question

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 ?

Answer 1

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}