3.67. 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.
3.67. By the condition, $A D \| B C, B C=2, \angle A B C=\angle B C D=135^{\circ}$, $\angle A O D=150^{\circ}$ (Fig. 3.69). Since the angles at the base of the trapezoid are equal, the trapezoid is isosceles, i.e., $A B=C D$. We find

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-543.jpg?height=375&width=624&top_left_y=101&top_left_x=42)

Fig. 3.69 $\angle A D C=180^{\circ}-135^{\circ}=45^{\circ} ;$

$\angle O D A=\frac{180^{\circ}-150^{\circ}}{2}=15^{\circ}$;

$\angle B D C=45^{\circ}-15^{\circ}=30^{\circ}$. From $\triangle B D C$ by the Law of Sines we have $\frac{B C}{\sin 30^{\circ}}=\frac{B D}{\sin 135^{\circ}}$, hence

$B D=\frac{2 \sin 45^{\circ}}{\sin 30^{\circ}}=2 \sqrt{2}$. Therefore,

$$
S_{A B C D}=\frac{1}{2} B D^{2} \sin 150^{\circ}=\frac{1}{2} \cdot 8 \cdot \frac{1}{2}=2
$$

(since the area of a convex quadrilateral is equal to $\frac{1}{2} d_{1} d_{2} \sin \alpha$, where $\alpha$ is the angle between the diagonals).