# 2007 AMC 12A Problems/Problem 10

The following problem is from both the 2007 AMC 12A #10 and 2007 AMC 10A #14, so both problems redirect to this page.

## 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 $\frac{4.8 \cdot 3.6}2 = 8.64\ \mathrm{(A)}$

## Solution 2

The hypotenuse of the triangle is a diameter of the circumcircle, so it has length $2 \cdot 3 = 6$. The triangle is similar to a 3-4-5 triangle with the ratio of their side lengths equal to $\frac{6}{5}$. The area of a 3-4-5 triangle is $\frac{3\cdot 4}{2} = 6$. The square of the ratio of their side lengths is equal to the ratio of their areas. Call the area of the triangle $A$. Therefore, $\left(\frac{6}{5}\right)^2 = \frac{A}{6} \Longrightarrow \frac{36}{25} = \frac{A}{6} \Longrightarrow A = \frac{36\cdot6}{25} = 8.64\ \mathrm{(A)}$

~mobius247

 2007 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
 2007 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions