how do i find the area of the shaded regionnnn?
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.