10.091. Find the area of a circle inscribed in an isosceles trapezoid, if its larger base is equal to $a$, and the angle at the smaller base is $120^{\circ}$.
Solution.

In $\triangle C E D, \angle C D E=60^{\circ}$ and $\cos 60=\frac{a-b}{2 c}$ (Fig. 10.88). Hence, $\frac{a-b}{2 c}=\frac{1}{2}$, $a-b=c$. For the circumscribed isosceles trapezoid, $a+b=2 c$.

Solving the system $\left\{\begin{array}{l}a-b=c, \\ a+b=2 c,\end{array}\right.$ we get $c=\frac{2}{3} a . \sin 60=\frac{2 R}{c}$. From this, $R=\frac{\sin 60 \cdot c}{2}=\frac{\sqrt{3}}{4} c=\frac{\sqrt{3}}{6} a$. The area of the circle $S=\pi R^{2}=\pi \frac{3}{36} a^{2}=\frac{\pi a^{2}}{12}$. Answer: $\frac{\pi a^{2}}{12}$.