12.194. In a trapezoid, the smaller base is equal to 2, the adjacent angles are $135^{\circ}$ each. The angle between the diagonals, facing the base, is $150^{\circ}$. Find the area of the trapezoid.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0942.jpg?height=243&width=679&top_left_y=918&top_left_x=595)

Fig. 12.61
## Solution.

Let in trapezoid $A B C D$, $B C=2, \angle A B C=\angle D C B=135^{\circ}, O$ - the point of intersection of the diagonals, $\angle B O C=150^{\circ}$ (Fig. 12.61).

In $\triangle B O C: \angle A C B=\angle D B C=15^{\circ}$, then $\angle B A C=180^{\circ}-(\angle A B C+$ $+\angle A C B)=30^{\circ}$.

By the Law of Sines in $\triangle A B C$ :

$$
\begin{gathered}
\frac{A C}{\sin \angle A B C}=\frac{B C}{\sin \angle B A C} \Rightarrow A C=\frac{B C \sin \angle A B C}{\sin \angle A B C}=\frac{2 \sin 135^{\circ}}{\sin 30^{\circ}}=2 \sqrt{2} \\
S_{\triangle A B C D}=\frac{1}{2} A C \cdot B D \sin \angle B O C=\frac{1}{2} A C^{2} \sin \angle B O C=\frac{1}{2}(2 \sqrt{2})^{2} \sin 150^{\circ}=2
\end{gathered}
$$

Answer: 2.