10.199. The bases of the trapezoid are 4 and 16 cm. Find the radii of the inscribed and circumscribed circles of the trapezoid, given that these circles exist.
Solution.

In trapezoid $ABCD$ (Fig. 10.9) $BC \| AD, BC=4$ cm, $AD=16$ cm. Since a circle is circumscribed around the given trapezoid, then $AB=CD$. Since a circle can be inscribed in the given trapezoid, then $AD+BC=AB+CD=2AB; AB=\frac{AD+BC}{2}=10$ cm. $BK$ is the height of the trapezoid. Then $AK=\frac{AD-BC}{2}=6$ cm. From $\triangle AKB \quad (\angle AKB=90^{\circ})$: $BK=\sqrt{AB^{2}-AK^{2}}=8$ cm. The radius of the inscribed circle $r=\frac{1}{2} BK=4$ cm.

The radius $R$ of the circumscribed circle will be found as the radius of the circle circumscribed around $\triangle ABD: R=\frac{AB \cdot BD \cdot AD}{4 S_{\triangle ABD}}, KD=AD-AK=10$ cm. From $\triangle BKD$ $\left(\angle BKD=90^{\circ}\right): BD=\sqrt{BK^{2}+KD^{2}}=2 \sqrt{41}$ cm. $S_{\triangle ABD}=\frac{1}{2} AD \cdot BK=64$ cm$^2$. Then $R=\frac{10 \cdot 2 \sqrt{41} \cdot 16}{4 \cdot 64}=\frac{5 \sqrt{41}}{4}$ (cm).

Answer: $4$ cm; $\frac{5 \sqrt{41}}{4}$ cm.