$19 \cdot 50$ As shown in the figure, $A F / / C D, A B / / E F, B C / /$ $E D$. Their lengths are all 1, and $\angle F A B=\angle B C D=$ $60^{\circ}$. What is the area of the figure?
(A) $\frac{\sqrt{3}}{2}$.
(B) 1 .
(C) $\frac{3}{2}$.
(D) $\sqrt{3}$.
(E) 2 .
(34th American High School Mathematics Examination, 1983)
[Solution] As shown in the figure, connect $B F$, $B E$, and $B D$, dividing the shape $A B C D E F$ into four triangles.

It is easy to verify that these four triangles are all equilateral triangles with side lengths of 1 (for example, in $\triangle F A B$, since $A F=A B=1$, we have
$$
\angle A F B=\angle A B F=\frac{180^{\circ}-60^{\circ}}{2}=60^{\circ} \text {. }
$$

Thus, the total area is
$$
4 \cdot \frac{1}{2} A F \cdot A B \cdot \sin 60^{\circ}=4 \cdot \frac{1^{2} \cdot \sqrt{3}}{4}=\sqrt{3} \text {. }
$$

Therefore, the answer is $(D)$.