10.148. Find the area of an isosceles trapezoid if its height is equal to $h$, and the lateral side is seen from the center of the circumscribed circle at an angle of $60^{\circ}$.
Solution.

Since the central angle $C O D$ is equal to $60^{\circ}$ (Fig. 10.132), the inscribed angle $C A D$ is equal to $30^{\circ}$. Therefore, $h=\frac{1}{2} A C$ and from $\triangle A K C$ we get $A K=\sqrt{A C^{2}-C K^{2}}=h \sqrt{3}$. We find the area of the trapezoid:

$$
S=\frac{1}{2}(B C+A D) h=(A E+E K) h=A K \cdot h=h \sqrt{3} \cdot h=h^{2} \sqrt{3}
$$

Answer: $\quad h^{2} \sqrt{3}$.