1. In trapezoid $A B C D$, $A D / / B C, E F$ is the midline, the ratio of the area of quadrilateral $A E F D$ to the area of quadrilateral $E B C F$ is $\frac{\sqrt{3}+1}{3-\sqrt{3}},$ and the area of $\triangle A B D$ is $\sqrt{3}$. Then the area of trapezoid $A B C D$ is $\qquad$
(2010, I Love Mathematics Junior High School Summer Camp Mathematics Competition)
Let $A D=a, B C=b, S_{\text {trapezoid } A B C D}=S$.
$$
\begin{array}{l}
\text { Then } E F=\frac{a+b}{2}, \frac{a}{b}=\frac{S_{\triangle A B D}}{S_{\triangle B C D}}=\frac{\sqrt{3}}{S-\sqrt{3}}, \\
\frac{S_{\text {quadrilateral } A E F D}}{S_{\text {quadrilateral } E B C F}}=\frac{a+\frac{a+b}{2}}{b+\frac{a+b}{2}}=\frac{3 a+b}{a+3 b} \\
=\frac{3 \sqrt{3}+S-\sqrt{3}}{\sqrt{3}+3 S-3 \sqrt{3}}=\frac{\sqrt{3}+1}{3-\sqrt{3}} .
\end{array}
$$

Therefore, $S=2$.