- Thread starter
- #1
Thanks I can for sure, we have the semicircle areaPerhaps this can get you started. Please refer to the following diagram:
View attachment 529
The area of the circular sector (the sum of the red and green areas) is:
$\displaystyle A_S = \frac{1}{2}r^2\theta$
Now, we see that:
$\displaystyle \cos(\theta)=\frac{k}{r}\,\therefore\,\theta=\cos^{-1}\left(\frac{k}{r} \right)$
and so we have:
$\displaystyle A_S = \frac{1}{2}r^2\cos^{-1}\left(\frac{k}{r} \right)$
The area of the green triangle is:
$\displaystyle A_T=\frac{1}{2}k\sqrt{r^2-k^2}$
And thus, the area A in red is:
$\displaystyle A=A_S-A_T=\frac{1}{2}\left(r^2\cos^{-1}\left(\frac{k}{r} \right)-k\sqrt{r^2-k^2} \right)$
Can you proceed from here?