10.054. Based on an isosceles triangle with a base equal to 8 cm, a circle is constructed on the base as a chord, touching the lateral sides of the triangle (Fig. 10.54). Find the radius of the circle if the length of the height drawn to the base of the triangle is 3 cm.
## Solution.

Given $A B=B Q, A Q=8 \text{ cm}, B E=3 \text{ cm}. A B=\sqrt{16+9}=5 \text{ cm}$,

$$
\begin{aligned}
& A E^{2}=B E \cdot O E \quad \Rightarrow \quad O E \cdot 3=16, \quad O E=\frac{16}{3} \cdot \quad A B^{2}=B D \cdot B C \\
& 25=\left(R+3+\frac{16}{3}\right) \cdot\left(\frac{25}{3}-R\right), R^{2}=\frac{625-25 \cdot 9}{9}=\frac{400}{9}, R=\frac{20}{3}
\end{aligned}
$$

Answer: $\quad \frac{20}{3} \text{ cm}$.