areacirclesgeometrytrigonometry
This is how it looks like:
It is given that the area of the shaded region is $35 cm^2$.
All of my attempts so far ended up in a two-variable equation in terms of $r_1$ and $r_2$ (the radii of the larger circle and smaller circle respectively).
So, how do I find the area enclosed between the two circles, that is, the area of the larger circle minus the area of the smaller circle?
Consider the image below:
Let $R$ be the radius of the big circle and $r$ the radius of the small circles.
If we connect the center of the small circles we get an equilateral triangle with side = $2r$. The center of the triangle is the center of the big circle as well.
The distance between the center of the triangle to the midpoint of one of its sides (aka apothem) is $h/3$, where $h$ is the height of the triangle. In our case, $h = r\sqrt{3}$ (we can easily calculate this with the pythagorean theorem).
As we can see in the image, we can now say that $R = r + \frac{2}{3}r\sqrt{3} = r \left(1 + \frac{2\sqrt{3}}{3}\right) \approx 2.1547r$.
$Area_{shaded} = 3\pi r^2 $
$Area_{bigcircle} = \pi R^2 = \pi (2.1547r)^2 \approx 4.6427\pi r^2$
$Area_{unshaded} = Area_{bigcircle} - Area_{shaded} \approx (4.6427 - 3) \pi r^2 \approx 1.6427 \pi r^2 $
Finally, we can check that $2 (Area_{unshaded}) \approx 3.2854 \pi r^2$ is bigger than $Area_{shaded} = 3 \pi r^2$
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}
Best Answer
Say $R_s$ is the radius of small circle and $R_b$ is the radius of big one, then
From (2) you just find, that $R_b^2 - R_s^2 = 70$ and then substituting it into (1) you get $70\pi$