10.213. The center of a semicircle inscribed in a right triangle, such that its diameter lies on the hypotenuse, divides the hypotenuse into segments of 30 and 40 cm. Find the length of the arc of the semicircle, enclosed between the points of tangency with the legs.
Solution.

Draw radii $O D$ and $O E$ to the points of tangency (Fig. 10.23). We have $O D=O E=C E=C D$, i.e., $E C D O$ is a square. Let $R$ be the radius of the circle; then the length of the arc $E D$ is $\frac{\pi R}{2}$. Since $\triangle A E O \sim \triangle O D B$,
we have $\frac{A E}{O D}=\frac{A O}{O B}=\frac{30}{40}=\frac{3}{4}$. But $A E^{2}=A O^{2}-O E^{2}=30^{2}-R^{2}$, from which $\frac{\sqrt{30^{2}-R^{2}}}{R}=\frac{3}{4}, 16\left(30^{2}-R^{2}\right)=9 R^{2} \Rightarrow R=24$. The length of the arc $E D$ is $12 \pi$. Answer: $12 \pi$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0725.jpg?height=301&width=717&top_left_y=411&top_left_x=348)

Fig. 10.24