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

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2007 AMC 12A Problems/Problem 10

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