10.099. The area of an isosceles trapezoid (Fig. 10.95) circumscribed about a circle is $8 \mathrm{~cm}^{2}$. Determine the sides of the trapezoid if the angle at the base is $30^{\circ}$.
Solution.

From $\triangle C E D$ we get $\sin 30^{\circ}=\frac{2 R}{c}$. Hence $2 R=\frac{c}{2}$. The area of the trapezoid $S=2 R c=\frac{c^{2}}{2}=8$. From this, $c^{2}=16, c=4$ (cm), $C D=A B=4$ (cm). We have $\cos 30^{\circ}=\frac{E D}{C D}, E D=2 \sqrt{3}$ (cm). Solving the system $\left\{\begin{array}{l}a+b=4, \\ b=a+4 \sqrt{3}\end{array}\right.$. From this, we get $a=4-2 \sqrt{3}, b=4+2 \sqrt{3}$.

Answer: $4-2 \sqrt{3} ; 4+2 \sqrt{3} ; 4$ and 4 cm.