4. Let $A B C$ be a triangle such that $A B=2, C A=3$, and $B C=4$. A semicircle with its diameter on $\overline{B C}$ is tangent to $\overline{A B}$ and $\overline{A C}$. Compute the area of the semicircle.
Answer: $\frac{27 \pi}{40}$
Solution: Let $O, D$, and $E$ be the midpoint of the diameter and the points of tangency with $\overline{A B}$ and $\overline{A C}$ respectively. Then $[A B C]=[A O B]+[A O C]=\frac{1}{2}(A B+$ $A C) r$, where $r$ is the radius of the semicircle. Now by Heron's formula, $[A B C]=$ $\sqrt{\frac{9}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{5}{2}}=\frac{3 \sqrt{15}}{4}$. We solve for $r=\frac{3 \sqrt{15}}{10}$ and compute $\frac{1}{2} \pi r^{2}=\frac{27 \pi}{40}$.