10.191. The center of the circle inscribed in a right trapezoid is at distances of 3 and 9 cm from the ends of its lateral side. Find the sides of the trapezoid.
Solution.

Given $O C=3$ cm, $O D=9$ cm (see Fig. 10.1). Let $N, P, M, E$ be the points of tangency of the circle with the sides of the trapezoid $B C, C D, A D$, $A B$ respectively. By the property of tangents:

1) $O N \perp B C, O P \perp C D, O M \perp A D$.
2) $\angle N C O=\angle P C O, \angle M D O=\angle P D O$.

Then $\angle O C D+\angle O D C=\frac{1}{2}(\angle B C D+\angle A D C)=90^{\circ} \Rightarrow \triangle C O D-$ is a right triangle, from which $C D=\sqrt{3^{2}+9^{2}}=3 \sqrt{10}$ cm. Let $r$ be the radius of the inscribed circle. $S_{\triangle C O D}=\frac{1}{2} C D \cdot O K=\frac{1}{2} O C \cdot O D$ or $3 \cdot \sqrt{10} \cdot r=3 \cdot 9$, $r=\frac{9}{\sqrt{10}}=\frac{9 \sqrt{10}}{10} \cdot$ From $\triangle C P O:$

$$
\begin{aligned}
& C P=\sqrt{O C^{2}-O P^{2}}=\sqrt{3^{2}-\frac{9^{2}}{10}}=3 \sqrt{1-\frac{9}{10}}=\frac{3}{\sqrt{10}}=\frac{3 \sqrt{10}}{10} \\
& P D=C D-C P=3 \sqrt{10}-\frac{3 \sqrt{10}}{10}=3 \sqrt{10}\left(1-\frac{1}{10}\right)=\frac{27 \sqrt{10}}{10}
\end{aligned}
$$

By the property of the tangent, $M D=P D, N C=C P$. Clearly, $B N O E$ is a square, so

$$
\begin{aligned}
& B E=B N=r=\frac{9 \sqrt{10}}{10} \cdot A B=2 r=\frac{9 \sqrt{10}}{5} \mathrm{~cm} \\
& B C=B N+C N=\frac{9 \sqrt{10}}{10}+\frac{3 \sqrt{10}}{10}=\frac{6 \sqrt{10}}{5} \mathrm{~cm} \\
& A D=A N+N D=\frac{9 \sqrt{10}}{10}+\frac{27 \sqrt{10}}{10}=\frac{18 \sqrt{10}}{5} \mathrm{~cm}
\end{aligned}
$$

Answer: $\frac{9 \sqrt{10}}{5}, \frac{6 \sqrt{10}}{5}, 3 \sqrt{10}, \frac{18 \sqrt{10}}{5} \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0711.jpg?height=422&width=567&top_left_y=78&top_left_x=94)

Fig. 10.2

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0711.jpg?height=389&width=586&top_left_y=96&top_left_x=676)

Fig. 10.3