Hello, razemsoft21!
Find the shaded area.
Code:* * * * |r/2 * * A| * * o * .|. * .:|:. * * .::O::. * * .::*|*::. * .::*:|:*::. * .::*::|::*::. * * ::*:::|:::*:: * Bo - - - - - oC r
$\displaystyle \,O$ is the center of the circle.
Draw $\displaystyle AB$ and $\displaystyle AC.$
Note that $\displaystyle \Delta OBC$ is equilateral.
Its height is: .$\displaystyle \frac{\sqrt{3}}{2}r$
The height of $\displaystyle \Delta ABC$ is: .$\displaystyle \frac{r}{2} + \frac{\sqrt{3}}{2}r \:=\:\left(\frac{1+\sqrt{3}}{2}\right)r$
The area of $\displaystyle \Delta ABC$ is: .$\displaystyle \frac{1}{2}bh \;=\;\frac{1}{2}(r)\left(\frac{1+\sqrt{3}}{2}\righ t)r \;=\;\dfrac{1+\sqrt{3}}{4}\,r^2 $
Ha! . . . That was stupid!
I found the wrong area . . . *blush*
Let's see, the shaded area is the area of the circle,
. . minus the area of $\displaystyle \Delta ABC$, minus that missing segment.
The area of the segment is the area of the $\displaystyle 60^o$ sector
. . minus equilateral triangle $\displaystyle OBC.$
The area of the sector is: .$\displaystyle \frac{1}{2}r^2(\frac{\pi}{3}) \:=\:\frac{\pi}{6}r^2$
The area of the equilateral triangle is: .$\displaystyle \frac{\sqrt{3}}{4}r^2 $
So the area of the segment is: .$\displaystyle \frac{\pi}{6}r^2 - \frac{\sqrt{3}}{2}r^2 \:=\:\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)r^2 $
Therefore, the desired area is:
. . $\displaystyle \displaystyle \pi r^2 - \left(\frac{1+\sqrt{3}}{4}\right)r^2 - \left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)r^2 \;=\;\left(\frac{5\pi}{6} - \frac{1}{4}\right)r^2$
Here is a slightly different method. In the attached picture (where the top of the white equilateral triangle is the centre of the circle), the blue and red regions together form 5/6ths of the area of the circle, so their combined area is $\displaystyle \tfrac56r^2$. From that we must subtract the red area. Each red triangle has area half the product of two sides times the sine of the included angle. In this case the sides are $\displaystyle \tfrac12r$ and $\displaystyle r$, and the included angle is $\displaystyle 150^\circ$. So the area of each triangle is $\displaystyle \tfrac14r^2\sin(150^\circ) = \tfrac18r^2$. Therefore the blue area is $\displaystyle \bigl(\tfrac{5\pi}6-\tfrac14\bigr)r^2$, as Soroban already showed.
Sometimes it's worth doing a problem in more than one way, just to check that you have the correct answer.