10.146. The length of the height dropped to the base of an isosceles triangle is 25 cm, and the radius of the inscribed circle is $8 \mathrm{~cm}$. Find the length of the base of the triangle.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-625.jpg?height=354&width=404&top_left_y=95&top_left_x=108)

Fig. 10.132
## Solution.

Let $K, L, M$ be the points of tangency (Fig. 131). Then $B K=25$ (cm), $O K=O L=O M=8$ (cm). Therefore,

$$
\begin{aligned}
& B O=B K-O K=17(\mathrm{~cm}) \\
& B M=\sqrt{B O^{2}-O M^{2}}=\sqrt{289-64}=15(\mathrm{~cm})
\end{aligned}
$$

Then $\operatorname{tg} \alpha=\frac{O M}{B M}=\frac{8}{15}$. On the other hand, $\operatorname{tg} \alpha=\frac{K C}{B K}$, from which $K C=B K \operatorname{tg} \alpha=25 \cdot \frac{8}{15}=\frac{40}{3}$ (cm). Therefore, $A C=2 K C=\frac{80}{3}(\mathrm{~cm})$.

Answer: $\frac{80}{3} \mathrm{~cm}$.