Suppose we have a triangle with a right angle at its height, with side a, 10 inches, side b, unknown, and side c, 24 inches; inscribed in a semi-circle. 
Now, I'm asked to find the area of the semi circle only, in terms of $\pi$. I was already given the answer, which is $84.5\pi -120 \ in^2$, but I don't understand how to get this conclusion.
I tried finding the area this way, but I get a pretty different answer:
$A=\frac{\pi r^2}{2}, r=12 \ inches$
$$A=\frac{144\pi}{2}$$
$$A=72\pi \ in^2$$
Then I know I have to subtract the triangle because I only want the area of the shaded region (the semi-circle), so
$$10^2+B^2=24^2$$
$$2\sqrt{119}$$
So I end up getting an answer of $A=72\pi-2\sqrt{119} \ in^2$ which is clearly false. Could anyone explain where I went wrong?
Best Answer
The question is poorly worded - it is not clear which side of the triangle is the longest (the hypotenuse).
You are assuming that the 24 inch side of the triangle is the longest side and is therefore the diameter of the semi circle.
The book answer is assuming that the unknown side of the triangle is the longest side. In this case the length of this side is $\sqrt{10^2+24^2}=26$ inches, the radius of the semicircle is $13$ inches, the area of the triangle is $\frac{10 \times 24}{2} = 120$ square inches and the area of the semicircle outside of the triangle is
$\frac{13^2\pi}{2}-120$ square inches.