The vertices of a $3-4-5$ right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles? $\mathrm{(A) \ } 12\pi\qquad$ $ \mathrm{(B) \ } \frac{25\pi}{2}\qquad$ $ \mathrm{(C) \ } 13\pi\qquad$ $ \mathrm{(D) \ } \frac{27\pi}{2}\qquad$ $ \mathrm{(E) \ } 14\pi$
Solution
Assume the three circles have radius $r_a, r_b, r_c$.
Then $r_a+r_b=3, r_a+r_c=4, r_b+r_c=5$, so we get $r_a=1, r_b=2, r_c=3$.
The vertices of a $5$-$12$-$13$ right triangle are the centers of three mutually externally tangent circles. What is the sum of the areas of the three circles?
$\mathrm{(A) \ } 12\pi\qquad$ $ \mathrm{(B) \ } \frac{25\pi}{2}\qquad$ $ \mathrm{(C) \ } 13\pi\qquad$ $ \mathrm{(D) \ } \frac{27\pi}{2}\qquad$ $ \mathrm{(E) \ } 14\pi$ 
