19・11 In a convex equiangular hexagon $A B C D E F$, $A B=1, B C=4, C D=2$ and $D E=4$. Then the area of the hexagon is
(A) $\frac{15}{2} \sqrt{3}$.
(B) $9 \sqrt{3}$.
(C) 16
(D) $\frac{39}{4} \sqrt{3}$.
(E) $\frac{43}{4} \sqrt{3}$.
(50th American High School Mathematics Examination, 1999)
[Solution] Connect $B E$. Given $B C=D E=4$, $\angle B C D=\angle C D E=120^{\circ}$, we know that quadrilateral $B C D E$ is an isosceles trapezoid, and $\angle C B E=\angle D E B=60^{\circ}$.
From this, we can deduce that $A F E B$ is also an isosceles trapezoid, so
$$
E F=1,
$$

Furthermore, since $\angle C B E=60^{\circ}$, we can calculate $B E=6$.
Similarly, we can calculate $A F=5$.
Thus, $S_{B C D E}=\frac{1}{2} \cdot(2+6) \cdot \sqrt{4^{2}-2^{2}}$
$$
=8 \sqrt{3},
$$

and $S_{A B E F}=\frac{1}{2} \cdot(5+6) \cdot \sqrt{1^{2}-\left(-\frac{1}{2}\right)^{2}}=\frac{11}{4} \sqrt{3}$.
$\therefore S_{A B C D E F}=8 \sqrt{3}+\frac{11}{4} \sqrt{3}=\frac{43}{4} \sqrt{3}$.
Therefore, the answer is $(E)$.