# Find the area of the four sectors of the given circle

#### MarkFL

Staff member
Perhaps this can get you started. Please refer to the following diagram: 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?

Last edited:

#### Amer

##### Active member
Perhaps 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?
Thanks I can for sure, we have the semicircle area
the quarter area of the circle minus the area of the red we will get the semi area of below sector.