10.186. The lengths of the bases of an isosceles trapezoid are in the ratio 5:12, and the length of its height is 17 cm. Calculate the radius of the circle circumscribed around the trapezoid, given that its midline is equal to the height.
Solution.

Let $B C=5 k, A D=12 k$ (Fig. 10.153). Then $\frac{B C+A D}{2}=h=17$, $\frac{12 k+5 k}{2}=17$, from which $k=2$. Therefore, $B C=10, A D=24$. Let $C L \perp A D$, $B K \perp A D, A K=L D=\frac{A D-B C}{2}=7$ (cm); $C D=\sqrt{C L^{2}+L D^{2}}=\sqrt{17^{2}+7^{2}}=$ $=13 \sqrt{2}, \sin \alpha=\frac{C L}{C D}=\frac{17}{13 \sqrt{2}}$. From $\triangle A L C$ we find $A C=\sqrt{C L^{2}+A L^{2}}=$ $=\sqrt{17^{2}+17^{2}}=17 \sqrt{2}$. By the Law of Sines $\frac{A C}{\sin \alpha}=2 R$, therefore, $\frac{17 \sqrt{2} \cdot 13 \sqrt{2}}{17 \cdot 2}=R, R=13(\text{cm})$.

Answer: 13 cm.