A [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) with side lengths in the [ratio](https://artofproblemsolving.com/wiki/index.php/Ratio) $3 : 4 : 5$ is inscribed in a [circle](https://artofproblemsolving.com/wiki/index.php/Circle) with [radius](https://artofproblemsolving.com/wiki/index.php/Radius) 3. What is the area of the triangle?
$\mathrm{(A)}\ 8.64\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 5\pi\qquad \mathrm{(D)}\ 17.28\qquad \mathrm{(E)}\ 18$
[2007 AMC12A-10.png](https://artofproblemsolving.com/wiki/index.php/File:2007_AMC12A-10.png)
Since 3-4-5 is a [Pythagorean triple](https://artofproblemsolving.com/wiki/index.php/Pythagorean_triple), the triangle is a [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle). Since the hypotenuse is a [diameter](https://artofproblemsolving.com/wiki/index.php/Diameter) of the [circumcircle](https://artofproblemsolving.com/wiki/index.php/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 $\frac{4.8 \cdot 3.6}2 = 8.64\ \mathrm{(A)}$