1993 AHSME Problems/Problem 14

Problem

$[asy] draw((-1,0)--(1,0)--(1+sqrt(2),sqrt(2))--(0,sqrt(2)+sqrt(13-2*sqrt(2)))--(-1-sqrt(2),sqrt(2))--cycle,black+linewidth(.75)); MP("A",(-1,0),SW);MP("B",(1,0),SE);MP("C",(1+sqrt(2),sqrt(2)),E);MP("D",(0,sqrt(2)+sqrt(13-2*sqrt(2))),N);MP("E",(-1-sqrt(2),sqrt(2)),W); dot((-1,0));dot((1,0));dot((1+sqrt(2),sqrt(2)));dot((-1-sqrt(2),sqrt(2)));dot((0,sqrt(2)+sqrt(13-2*sqrt(2)))); [/asy]$

The convex pentagon $ABCDE$ has $\angle{A}=\angle{B}=120^\circ,EA=AB=BC=2$ and $CD=DE=4$ . What is the area of ABCDE?

$\text{(A) } 10\quad \text{(B) } 7\sqrt{3}\quad \text{(C) } 15\quad \text{(D) } 9\sqrt{3}\quad \text{(E) } 12\sqrt{5}$

Solution

$\fbox{B}$

1993 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1993 AHSME Problems/Problem 14

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