One of the bases of the trapezoid serves as the diameter of a circle with radius $R$, while the other base is a chord that cuts off an arc of $\alpha$ radians from the circle ( $0<\alpha<\pi$ ). Find the area of the trapezoid.
The given trapezoid is isosceles.

## Solution

Let $A D$ be the diameter of the circle circumscribed around trapezoid $A B C D$, and $O$ be the center of this circle (the midpoint of $A D$). Then $\angle B O C = \alpha$.

Let $O M$ be the height of the trapezoid. Then

$$
\angle M O C = \frac{\alpha}{2}, \quad O M = R \cos \frac{\alpha}{2}, \quad M C = O C \sin \frac{\alpha}{2} = R \sin \frac{\alpha}{2}, \quad B C = 2 M C = 2 R \sin \frac{\alpha}{2}
$$

Therefore,

$$
S_{\mathrm{ABCD}} = \frac{1}{2}(A D + B C) \cdot O M = \left(R + R \sin \frac{\alpha}{2}\right) R \cos \frac{\alpha}{2} = R^{2}\left(1 + \sin \frac{\alpha}{2}\right) \cos \frac{\alpha}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-10.jpg?height=406&width=489&top_left_y=512&top_left_x=790)

## Answer

$R^{2}\left(1 + \sin \frac{\alpha}{2}\right) \cos \frac{\alpha}{2}$.