10.096. The area of an isosceles trapezoid circumscribed around a circle is $32 \sqrt{3} \mathrm{~cm}^{2}$ (Fig. 10.92). Determine the lateral side of the trapezoid, given that the acute angle at the base is $\pi / 3$.
Solution.

$\triangle C E D-$ is a right triangle and $\sin 60^{\circ}=\frac{2 R}{C D}$. Therefore, $2 R=\frac{\sqrt{3}}{2} C D$, $C D=x \Rightarrow 2 R=\frac{\sqrt{3}}{2} x$. The area of the trapezoid $S=\frac{a+b}{2} h=2 R x=\frac{\sqrt{3}}{2} x^{2}$. According to the condition $\frac{\sqrt{3}}{2} x^{2}=32 \sqrt{3} \Rightarrow x^{2}=64, x=8$ (cm).

Answer: 8 cm.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-602.jpg?height=348&width=624&top_left_y=112&top_left_x=96)

Fig. 10.93

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-602.jpg?height=390&width=518&top_left_y=75&top_left_x=733)

Fig. 10.94