10.408 An isosceles trapezoid is circumscribed around a circle with a radius of 5 cm. The distance between the points of tangency of its lateral sides is 8 cm. Find the area of the trapezoid.
Solution.

Let $O$ be the center of the circle with radius 5, around which an isosceles trapezoid $ABCD$ is circumscribed, and $N, L, M, K$ are the points of tangency with the circle.

$$
\begin{gathered}
B M=B N=C N=C L(\text { theorem } 3) \Rightarrow \\
\Rightarrow M L \| B C(\text { Thales' theorem }) \Rightarrow \\
\Rightarrow M L \perp N K \Rightarrow S_{M N L K}=\frac{M L \cdot N K}{2}=40 \\
S_{M N K}=\frac{S_{M N K L}}{2}=20
\end{gathered}
$$

$\angle B A O=\angle O A K=\angle M K O$, since $M K \perp A O$ ( $A O$ is the bisector and, therefore, the height of the isosceles $\triangle A M K$ ). Additionally, $\angle N M K=90^{\circ}$, as it subtends the diameter $N K$ (theorem 6). $\angle A O B=90^{\circ}$ (theorem 9). Therefore,

$$
\begin{gathered}
\triangle M N K \sim \triangle A O B \Rightarrow S_{A O B}=20\left(\frac{A B}{N K}\right)^{2}(\text { theorem 7) }= \\
=\frac{A B \cdot O M}{2} \Rightarrow A B=\frac{25}{2}(O M=5, N K=10) \Rightarrow \frac{25}{2}=\frac{B C+A D}{2}(\text { theorem 2) } \Rightarrow \\
\Rightarrow S_{A B C D}=10 \cdot \frac{B C+A D}{2}=125 .
\end{gathered}
$$

Answer: $125 \text{ cm}^2$.