# 1983 AHSME Problems/Problem 4

## Problem 4

In the adjoining plane figure, sides $AF$ and $CD$ are parallel, as are sides $AB$ and $EF$, and sides $BC$ and $ED$. Each side has length $1$. Also, $\angle FAB = \angle BCD = 60^\circ$. The area of the figure is $\textbf{(A)} \ \frac{\sqrt 3}{2} \qquad \textbf{(B)} \ 1 \qquad \textbf{(C)} \ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3}\qquad \textbf{(E)}\ 2$

## Solution $[asy] pair A, B, C, D, E, F; A = (0, 1.732); B = (0.5, 0.866); C = (0,0); D = (1, 0); E = (1.5, 0.866); F = (1, 1.732); draw(A--B--C--D--E--F--A); label("A", A, NW); label("B", B, 3W); label("C", C, SW); label("D", D, SE); label("E", E, E); label("F", F, NE); draw(B--D, dashed+linewidth(0.5)); draw(B--E, dashed+linewidth(0.5)); draw(B--F, dashed+linewidth(0.5)); [/asy]$

By rotating the diagram and drawing the dotted lines, we see that the figure can be divided into four equilateral triangles, each of side length $1$. The area of one such equilateral triangle is $\frac{\sqrt{3}}{4} \cdot 1^2 = \frac{\sqrt{3}}{4}$, which gives a total of $4\left(\frac{\sqrt{3}}{4}\right) = \sqrt{3}$, or $\boxed{D}$.

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