1.64. Find the radius of the circle circumscribed around an isosceles trapezoid with bases 2 and 14 and a lateral side of 10.
1.64. Note that the circle circumscribed around trapezoid $ABCD$ is also circumscribed around $\triangle ACD$ (Fig. 1.50), and such a circle is unique. We will find its radius using the formula $R=\frac{abc}{4S}$, where $S=S_{\triangle ACD}$. 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 $\triangle ACM$, we find $AC = \sqrt{AM^2 + MC^2} = \sqrt{8^2 + 8^2} = 8\sqrt{2}$. Finally, we get

$$
R = \frac{14 \cdot 10 \cdot 8\sqrt{2}}{4 \cdot 56} = 5\sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-443.jpg?height=636&width=483&top_left_y=726&top_left_x=41)

Fig. 1.51
Answer: $5\sqrt{2}$.