## Solution.

By the tangent-secant theorem, we have (Fig. 10.37) $B K^{2}=A B \cdot D B=9 \cdot 4=36$, from which $B K=6 \text{ cm}, K C=3 \text{ cm}$. We draw the radius $O K$ to the point of tangency and $O N \perp A B$. Then $A N=N D=\frac{5}{2} \text{ cm}$. Since $\triangle A N O \sim \triangle O K C$ (right triangles with $\angle A=\angle C$), we have $\frac{K C}{A N}=\frac{O C}{A O}=\frac{3}{5 / 2}=\frac{6}{5}$. Let the desired radius be $r$. Then $A O=r$, $O C=\sqrt{9+r^{2}}$, and thus $6 r=5 \sqrt{9+r^{2}}, 36 r^{2}=225+25 r^{2}$, from which $r=\frac{15}{\sqrt{11}} \text{ cm}$. Answer: $\frac{15}{\sqrt{11}} \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0734.jpg?height=467&width=731&top_left_y=51&top_left_x=321)

Fig. 10.38
10.228. One end of the diameter of a semicircle coincides with the vertex of the angle at the base of an isosceles triangle, and the other lies on this base. Find the radius of the semicircle if it touches one lateral side and divides the other into segments of lengths 5 and 4 cm, measured from the base.