# Difference between revisions of "2007 AMC 12A Problems/Problem 10"

## Problem

A triangle with side lengths in the ratio $3 : 4 : 5$ is inscribed in a circle with radius 3. What is the area of the triangle?

$\mathrm{(A)}\ 8.64\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 5\pi\qquad \mathrm{(D)}\ 17.28\qquad \mathrm{(E)}\ 18$

## Solution

Since 3-4-5 is a Pythagorean triple, the triangle is a right triangle. Since the hypotenuse is a diameter of the circumcircle, the hypotenuse is $2r = 6$. Then the other legs are $\frac{24}5=4.8$ and $\frac{18}5=3.6$. The area is $\frac{4.8 \cdot 3.6}2 = 8.64\ \mathrm{(A)}$