What is the radius of the circle inscribed in triangle $ABC$ if $AB = 15,$ $AC = 41,$ and $BC = 52$?
+1224 A = rs
semiperimeter = s = (15 + 41 + 52)/2 = 54
area = A = \(\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(54)(2)(39)(13)} = 234\)
radius = r = \(\frac{A}{s} = \frac{234}{54} = \frac{13}{3}\)
