1983 AHSME Problems/Problem 4

Problem 4

Position $A,B,C,D,E,F$ such that $AF$ and $CD$ are parallel, as are sides $AB$ and $EF$ , and sides $BC$ and $ED$ . Each side has length of $1$ and it is given that $\measuredangle FAB = \measuredangle BCD = 60^\circ$ . The area of the figure is

$\text{(A)} \ \frac{\sqrt 3}{2} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ \frac{3}{2} \qquad \text{(D)}\ \sqrt{3}\qquad \text{(E)}$

Solution

$[asy] pair A, B, C, D, E, F; A = (0, 1.732); B = (0.5, 0.866); C = (0,0); D = (1, 0); E = (1.5, 0.866); F = (1, 1.732); draw(A--B--C--D--E--F--A); label("A", A, A); label("B", (0.3, 0.866)); label("C", (-0.1, 0)); label("D", D, D); label("E", E, E); label("F", F, F); draw(B--D, dashed+linewidth(0.5)); draw(B--E, dashed+linewidth(0.5)); draw(B--F, dashed+linewidth(0.5)); [/asy]$

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)
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1983 AHSME Problems/Problem 4

Problem 4

Solution

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