$$
\begin{array}{r}
19 \cdot 4 \text { In pentagon } A B C D E, \angle A=\angle B \\
=120^{\circ}, E A=A B=B C=2, C D=D E=4 .
\end{array}
$$
The area of $A B C D E$ is
(A) 10 .
(B) $7 \sqrt{3}$.
(C) 15 .
(D) $9 \sqrt{3}$.
(44th American High School Mathematics Examination, 1993)
[Solution] Connect $C E$, it is easy to know that $A B C E$ is an isosceles trapezoid. Let $F, G$ be the feet of the perpendiculars from $A, B$ to $C E$, respectively, then $E F=C G$.
Since $\angle G B C=30^{\circ}$, we have $C G=\frac{1}{2} B C=1, B G=\sqrt{3}$.
And $C E=C G+G F+F E=1+2+1=4$, so $\triangle C D E$ is an equilateral triangle. Then
$$
\begin{aligned}
S_{A B C E} & =\frac{1}{2} B G \cdot(A B+C E) \\
& =\frac{1}{2} \sqrt{3}(2+4)=3 \sqrt{3}, \\
S_{\triangle C D E} & =\frac{\sqrt{3}}{4} C E^{2}=\frac{\sqrt{3}}{4} \cdot 16=4 \sqrt{3} .
\end{aligned}
$$

Therefore, $S_{A B C D E}=S_{A B C E}+S_{\triangle C D E}=7 \sqrt{3}$.
Hence, the answer is $(B)$.