10.072. In an isosceles triangle, the base is $30 \mathrm{~cm}$, and the lateral side is 39 cm. Determine the radius of the inscribed circle.
Solution.

Draw $B D \perp A C$ (Fig. 10.70); since $\triangle A B C$ is isosceles, $B D$ is also a median. We have $B D^{2}=A B^{2}-A D^{2}$, from which $B D=\sqrt{39^{2}-15^{2}}=36$ (cm) and, therefore, $S=\frac{1}{2} \cdot 30 \cdot 36=540\left(\mathrm{~cm}^{2}\right)$. But $S=p r=$ $=54 r$, from which $r=10 \mathrm{~cm}$.

. Answer: $10 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-590.jpg?height=494&width=520&top_left_y=81&top_left_x=124)

Fig. 10.71

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-590.jpg?height=464&width=436&top_left_y=108&top_left_x=782)

Fig. 10.72