$3+$ [ Law of Sines

In an isosceles trapezoid, the bases are given as $a=21, b=9$ and the height $h=8$. Find the radius of the circumscribed circle.

#
A circle circumscribed around trapezoid $ABCD$ is also circumscribed around triangle $ACD$.

## Solution

From vertex $C$ of the smaller base $BC$ of trapezoid $ABCD$, drop a perpendicular $CK$ to the larger base $AD$. Then $CK=8$. If $AD=21, BC=9$, then

\[
\begin{gathered}
KD=\frac{AD-BC}{2}=6, CD=\sqrt{CK^{2}+DK^{2}}=\sqrt{8^{2}+6^{2}}=10 \\
\sin \angle D=\frac{CK}{CD}=\frac{4}{5}, AC=\sqrt{AK^{2}+CK^{2}}=\sqrt{15^{2}+8^{2}}=17
\end{gathered}
\]

If $R$ is the radius of the circle circumscribed around trapezoid $ABCD$, then

\[
R=\frac{AC}{2 \sin \angle D}=\frac{85}{8}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-55.jpg?height=443&width=504&top_left_y=2271&top_left_x=777)

Answer

Find the radius of the circle circumscribed around an isosceles trapezoid with bases 2 and 14 and a lateral side of 10.

## Answer

$5 \sqrt{2}$.