1. As shown in Figure 3, in trapezoid $A B C D$, $A B / / C D$, $A B=a, C D=b(a$ $<b), \angle B C D=$ $60^{\circ}, \angle A D C=45^{\circ}$. The area of trapezoid $A B C D$ is equal to $\qquad$ (express the area in terms of $a$ and $b$).
$$
\text { II. 1. } \frac{3-\sqrt{3}}{4}\left(b^{2}-a^{2}\right) \text {. }
$$

As shown in Figure 8, construct $A E \perp$
$$
D C, B F \perp D C, E \text{ and } F
$$

as the feet of the perpendiculars. Then
$$
\begin{array}{l}
E F=A B=a . \\
\text { Let } A E=B F=h,
\end{array}
$$

then $D E=A E=h$.
$$
C F=\frac{B F}{\tan 60^{\circ}}=\frac{h}{\sqrt{3}} .
$$

Also, $b=C D=E F+D E+C F=a+h+\frac{h}{\sqrt{3}}$,
thus, $h=\frac{\sqrt{3}(b-a)}{1+\sqrt{3}}=\frac{3-\sqrt{3}}{2}(b-a)$.