1993 AMC 12 第 14 题

Draw $\overline{CE}$ to split the pentagon into a trapezoid and a triangle

The angle conditions give $CE=4,$ so $\triangle CDE$ is equilateral

Place $A=(0,0)$ and $B=(2,0).$ The $120^\circ$ angles and side lengths $EA=BC=2$ give $$E=(-1,\sqrt3),\qquad C=(3,\sqrt3).$$ Thus $ABCE$ is a trapezoid of height $\sqrt3$ and bases $2$ and $4,$ so its area is $3\sqrt3.$ Also $CE=4=CD=DE,$ making $\triangle CDE$ equilateral with area $4\sqrt3.$ The total area is $7\sqrt3.$ Thus the correct answer is B.