10.174. The area of an isosceles trapezoid circumscribed around a circle is $S$. Determine the radius of this circle if the angle at the base of the trapezoid is $30^{\circ}$.
## Solution.

Let $B C \| A D, \angle A=30^{\circ}$ (Fig. 10.144). Denote $B C=a, A D=b$;

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-637.jpg?height=226&width=530&top_left_y=191&top_left_x=111)

Fig. 10.145

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-637.jpg?height=320&width=520&top_left_y=94&top_left_x=716)

Fig. 10.146

Drop a perpendicular $B K$ to $A D$, denote it as $h, A B=c$. Since the circle is inscribed in the trapezoid, we have $a+b=2 c$ or $c=\frac{a+b}{2}$. From $\triangle A B K$, we find $\sin 30^{\circ}=\frac{h}{c}=\frac{2 h}{a+b}, \frac{2 h}{a+b}=\frac{1}{2}$, from which $a+b=4 h$. Therefore, $S=\frac{a+b}{2} \cdot h=2 h^{2}$; since $h=2 r$, we have $S=8 r^{2}$, from which $r=\sqrt{\frac{S}{8}}=\frac{\sqrt{2 S}}{4}$. Answer: $\frac{\sqrt{2 S}}{4}$