12.027. A trapezoid with angles $\alpha$ and $\beta$ at the larger base is circumscribed around a circle of radius $R$. Find the area of this trapezoid.
Solution.

Let $\angle BAD = \alpha, \angle CDA = \beta, OE = R$ (Fig. 12.30). Then $S_{ABCD} = \frac{BC + AD}{2} \cdot 2R \cdot S_{ABCD} = (BC + AD) \cdot R \cdot$ Since $BC + AD = AB + CD$, we have $BC + AD = 2R\left(\frac{1}{\sin \alpha} + \frac{1}{\sin \beta}\right)\left(AB = \frac{2R}{\sin \alpha}, CD = \frac{2R}{\sin \beta}\right)$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-735.jpg?height=322&width=502&top_left_y=95&top_left_x=137)

Fig. 12.30

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-735.jpg?height=316&width=530&top_left_y=102&top_left_x=685)

Fig. 12.31

\[
\begin{aligned}
& S_{ABCD} = 2R^2\left(\frac{1}{\sin \alpha} + \frac{1}{\sin \beta}\right) = \frac{2R^2(\sin \alpha + \sin \beta)}{\sin \alpha \sin \beta} = \frac{4R^2 \cdot \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}}{\sin \alpha \sin \beta} \\
& \text{Answer: } \frac{4R^2 \cdot \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}}{\sin \alpha \sin \beta}
\end{aligned}
\]