10.309. The area of an isosceles trapezoid circumscribed about a circle is $32 \mathrm{~cm}^{2}$; the acute angle of the trapezoid is $30^{\circ}$. Determine the sides of the trapezoid.
## Solution.

Let the height $B H$ of the trapezoid be $h$ (Fig. 10.98). Then $A B=2 h$ (since $\angle A=30^{\circ}$), $B C+A D=4 h$ (since $B C+A D=A B+C D=2 A B$). The area of the trapezoid $S=\frac{(B C+A D) h}{2}=2 h^{2}=32\left(\mathrm{~cm}^{2}\right)$, from which $h=4$ cm. Therefore, $A B=C D=8 \mathrm{~cm}, B C+A D=16 \mathrm{~cm}$. But $B C+A D=2 B C+2 A H=$ $=2 B C+8 \sqrt{3}$. Thus, $A B=C D=8 \mathrm{~cm}, B C=8-4 \sqrt{3} \mathrm{~cm}, A D=8+4 \sqrt{3} \mathrm{~cm}$.

Answer: 8 cm; $8-4 \sqrt{3}$ cm; $8+4 \sqrt{3}$ cm.