10.040. A semicircle is inscribed in a right triangle such that the diameter lies on the hypotenuse, and the center divides the hypotenuse into segments of lengths 15 and 20 cm (Fig. 10.40). Find the area of the triangle and the length of the inscribed semicircle.
Solution.

Given $A O=20 \mathrm{~cm}, O B=15 \mathrm{~cm}$. Let $r$ be the radius of the semicircle; $\triangle A F O \sim \triangle O E B$, we have $\frac{20}{15}=\frac{r}{\sqrt{225-r^{2}}} ; 4 \sqrt{225-r^{2}}=3 r ;$ $16\left(225-r^{2}\right)=9 r^{2}, r^{2}=9 \cdot 16, r=12 \mathrm{~cm} ; l=\pi r=12 \pi$. In $\triangle A F O, \angle A F O=90^{\circ}$, $A C=\sqrt{400-144}+12=28, B C=\sqrt{225-144}+12=21, S=\frac{28 \cdot 21}{2}=294 \mathrm{~cm}^{2}$.

Answer: $294 \mathrm{~cm}^{2}, 12 \pi \mathrm{~cm}$.