10.210. Find the radius of the circle circumscribed around an isosceles trapezoid with bases 2 and 14 and a lateral side of $10 \mathrm{~cm}$.
Solution.

The circle circumscribed around trapezoid $ABCD$ is also circumscribed around $\triangle ACD$ (Fig. 10.20), and such a circle is unique. Its radius is found using the formula $R=\frac{abc}{4S}$. We have $S=\frac{1}{2} AD \cdot BE$, where $BE=\sqrt{AB^2 - AE^2} = \sqrt{10^2 - 6^2} = 8$; hence, $S_{\triangle ACD} = \frac{1}{2} \cdot 14 \cdot 8 = 56$. From

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0723.jpg?height=466&width=458&top_left_y=94&top_left_x=203)

Fig. 10.20

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0723.jpg?height=500&width=424&top_left_y=63&top_left_x=741)

Fig. 10.21

$\triangle ACM$ we have $AC = \sqrt{AM^2 + MC^2} = \sqrt{8^2 + 8^2} = 8\sqrt{2}$. We obtained $R = \frac{14 \cdot 10 \cdot 8\sqrt{2}}{4 \cdot 56} = 5\sqrt{2}$.

Answer: $5\sqrt{2}$.