10.009. In an isosceles triangle, the base is 16 cm, and the lateral side is $10 \mathrm{~cm}$. Find the radii of the inscribed and circumscribed circles and the distance between their centers.
Solution.

$$
S_{\triangle A B C}=\frac{1}{2} A C \cdot B D, \text { where } B D=\sqrt{A B^{2}-A D^{2}}=\sqrt{10^{2}-8^{2}}=6 \text { (Fig.10.9), }
$$

i.e., $S_{\triangle A B C}=6 \cdot 8=48\left(\mathrm{~cm}^{2}\right)$. Let $R$ and $r$ be the radii of the circumscribed and inscribed circles of the triangle. Then

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-560.jpg?height=418&width=502&top_left_y=91&top_left_x=125)

Fig. 10.9

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-560.jpg?height=418&width=500&top_left_y=91&top_left_x=734)

Fig. 10.10

$$
R=\frac{A B \cdot B C \cdot A C}{4 S}=\frac{10 \cdot 10 \cdot 16}{4 \cdot 48}=\frac{25}{3}(\mathrm{~cm}), r=\frac{S}{p}=\frac{48}{18}=\frac{8}{3}(\mathrm{~cm}),
$$

where $p$ is the semiperimeter of $\triangle A B C$. The distance between the centers

$$
\begin{aligned}
& O_{1} O_{2}=O_{2} B-O_{1} B=O_{2} B-\left(B D-O_{1} D\right)=R-(6-r)=R+r-6= \\
& \quad=\frac{25}{3}+\frac{8}{3}-6=11-6=5(\mathrm{~cm}) .
\end{aligned}
$$

Answer: $\frac{8}{3}, \frac{25}{3}$ and 5 cm.