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
A $3k$-$4k$-$5k$ triangle is a right triangle (since $3^2+4^2=5^2$), with the right angle opposite the hypotenuse $5k$, which is therefore a diameter of the circumscribed circle.
$5k=2R=6$, so $k=\frac65$; the legs are $\frac{18}5$ and $\frac{24}5$.
Area $=\frac12\cdot\frac{18}5\cdot\frac{24}5=\frac{216}{25}=8.64$.
