A triangle has side lengths of 10, 24, and 26. Let a be the area of the circumcircle, and let b be the area of the triangle. Compute a - b in terms of pi.
+363 My best shot:
Okay so i don't know how to explain this, but the hypotenuse of this triangle is the diameter of the circle, so the circle area is
\(\pi\times r^2\)
\(169\pi\)
the area of the triangle is
\(\frac{10*24}{2}\)
\(120\)
and $s = (10+24+26)/2 = 30$, so $r = [ABC]/s = 120/30 = 4$.
When $[ABC] = rs$, where $r$ is the inradius and $s$ is the semiperimeter of the triangle.
Therefore, the area of the incircle is $4^2\pi = 16\pi$.Finally, the area of the circumcircle is $169\pi - 16\pi = \(\boxed{153\pi}$\)
