The lateral side of an isosceles trapezoid is equal to $a$, the midline is equal to $b$, and one angle at the larger base is $30^{\circ}$. Find the radius of the circle circumscribed around the trapezoid.
The projection of the diagonal of an isosceles trapezoid onto the larger base is equal to the midline of the trapezoid.

## Solution

Let $F$ be the projection of vertex $C$ of the smaller base $BC$ of the isosceles trapezoid $ABCD$ onto the larger base $AD$. Then the segment $AF$ is equal to the midline of the trapezoid, and since in the right triangle $CFD$ the angle $D$ is $30^{\circ}$, we have $CF = \frac{1}{2} CD$. From the right triangle $ACF$, we find that

\[
AC = \sqrt{AF^2 + CF^2} = \sqrt{b^2 + \frac{a^2}{4}}
\]

If $R$ is the radius of the circle circumscribed around triangle $ACD$, then

\[
R = \frac{AC}{2 \sin \angle D} = \sqrt{b^2 + \frac{a^2}{4}}
\]

It remains to note that the circle circumscribed around triangle $ACD$ coincides with the circle circumscribed around trapezoid $ABCD$.

## Answer

\(\sqrt{b^2 + \frac{a^2}{4}}\).