# Difference between revisions of "2006 iTest Problems/Problem 11"

## Problem

Find the radius of the inscribed circle of a triangle with sides of length $13$, $30$, and $37$.

$\text{(A) }\frac{9}{2}\qquad \text{(B) }\frac{7}{2}\qquad \text{(C) }4\qquad \text{(D) }-\sqrt{2}\qquad \text{(E) }4\sqrt{5}\qquad \text{(F) }6\qquad \\ \text{(G) }\frac{11}{2}\qquad \text{(H) }\frac{13}{2}\qquad \text{(I) }\text{none of the above}\qquad \text{(J) }1\qquad \text{(K) }\text{no triangle exists}\qquad$

## Solution

Since we know three sides, we can calculate the semiperimeter and area and use the results to find the radius of the incircle. The semiperimeter of the triangle is $\tfrac12 \cdot (13 + 30 + 37) = 40$. By Heron's Formula, the area of the triangle is $\sqrt{40 \cdot 27 \cdot 10 \cdot 3} = 180$. Since the radius of the incircle times the semiperimeter equals the area, $A = sr$ the radius of the incircle must be $\tfrac{180}{40} = \boxed{\textbf{(A) }\tfrac{9}{2}}$.