Area of Semi Circle With an Inscribed Triangle

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?

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.