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$", F, NE); draw(B--D, dashed+linewidth(0.5)); draw(B--E, dashed+linewidth(0.5)); draw(B--F, dashed+linewidth(0.5)); [/asy]$

Drawing the diagram as described, we create a convex hexagon with all side lengths equal to 1. In this case, it is a natural approach to divide the figure up into four equilateral triangles. The area, A, of one such equilateral triangle is $\frac{\sqrt{3}}{4}$ , which gives a total of $4\left(\frac{\sqrt{3}}{4}\right) = \sqrt{3}$ , or $\boxed{D}$ .

1983 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1983 AHSME Problems/Problem 4

Problem 4

Solution

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