# Thread: area in a circle?

1. ## area in a circle?

how do i find the area of the shaded regionnnn?

2. Originally Posted by hellothere

how do i find the area of the shaded regionnnn?
Is anything else known about the triangle?

3. I'm assuming the triangle is equilateral. You should be able to find the radius of the circle HA easily, since it's part of a 30-60-90 triangle. Then you can find the area of the the circle, subtract the area of the triangle, and divide by 3.

4. Assuming the triangle $\displaystyle {\Delta}ABC$ is equilateral we can find the radius by finding the length of $\displaystyle HA$.
We can create two $\displaystyle 30/60/90$ triangles within $\displaystyle {\Delta}ABC:{\Delta}AHD$ and $\displaystyle {\Delta}HCD$, which forms $\displaystyle {\Delta}AHC$.
We know $\displaystyle {\angle}DAH$ is equal to $\displaystyle 30$. Thus, we can let $\displaystyle sin(30)=\frac{5}{h}$ and solve for $\displaystyle h$. Doing so, we get $\displaystyle h=\frac{5}{sin(30)}=10$.
Now, we use Pythagoreans theorem to find the radius (length $\displaystyle HA$),
$\displaystyle 10^2=5^2+d^2 \longrightarrow d^2=10^2-5^2 \longrightarrow d=5\sqrt{3} \longrightarrow 2d=10\sqrt{3}$
So, the area of $\displaystyle {\Delta}AHC=\frac{1}{2}(10\sqrt{3})(5)=25\sqrt{3}$
Since we know $\displaystyle {\Delta}AHD$ and $\displaystyle {\Delta}HCD$ are both $\displaystyle 30/60/90$ triangles, we know $\displaystyle {\Delta}AHC$ gives us $\displaystyle {\angle}AHC=120$.
Therefore, the area of the sector $\displaystyle HADF=\frac{120}{360}\pi(10^2)=\frac{100\pi}{3}$.
Finally, we just subtract the area of the sector from the area of $\displaystyle {\Delta}AHC$ to get $\displaystyle \frac{100\pi}{3}-25\sqrt{3}=61.41$

Sorry, I shoulda drawn a picture...may be confusing.

5. Correct.

Edit. Oops. I thought you were the original poster.