10.171. The bases of the trapezoid are equal to $a$ and $b$, and the angles at the larger base are $\pi / 6$ and $\pi / 4$. Find the area of the trapezoid.
## Solution.

Let $B C \| A D$ and $\angle A=\pi / 6, \angle D=\pi / 4$. Drop perpendiculars $B K$ and $C L$ to the side $A D$ (Fig. 10.142). Let $C L=B K$ be denoted by $h$. Then $\operatorname{tg} \pi / 4=\frac{C L}{L D}$. Therefore, $L D=h$. Then $A K=b-a-h$, and $\operatorname{tg} \pi / 6=\frac{B K}{A K}=$ $=\frac{h}{b-a-h}, \frac{1}{\sqrt{3}}=\frac{h}{b-a-h}$, from which $h=\frac{b-a}{\sqrt{3}+1}$. The area of the trapezoid $S=\frac{a+b}{2} h=\frac{a+b}{2} \cdot \frac{b-a}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)\left(b^{2}-a^{2}\right)}{4}$.

Answer: $\frac{(\sqrt{3}-1)\left(b^{2}-a^{2}\right)}{4}$.