1.33. Since the central angle $C O D$ is $60^{\circ}$ (Fig. 1.29), the inscribed angle $C A D$ is $30^{\circ}$. Therefore, $h=\frac{1}{2} A C$ and from

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-428.jpg?height=467&width=517&top_left_y=1151&top_left_x=303)

Fig. 1.29

$\triangle A K C$ 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: $h^{2} \sqrt{3}$.
1.33. 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 $60^{\circ}$.