# 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

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

First, drop perpendiculars from points $A$ and $B$ to segment $EC$.

Since $\angle EAB = 120^{\circ}$, $\angle EAF = 30^{\circ}$.

This implies that $\triangle EAF$ is a 30-60-90 Triangle, so $EF = 1$ and $AF = \sqrt3$.

Similarly, $GB = \sqrt3$ and $GC = 1$.

Since $FABG$ is a rectangle, $FG = AB = 2$.

Now, notice that since $EC = EF + FG + GC = 1+2+1=4$, triangle $DEC$ is equilateral.

Thus, $[ABCDE] = [EABC]+[DCE] = \frac{2+4}{2}(\sqrt3)+\frac{4^2\cdot\sqrt3}{4} = 3\sqrt3+4\sqrt3=\boxed{7\sqrt3 (B)}$

-AOPS81619