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$", A, NW); label("$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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1983 AHSME Problems/Problem 4

Problem 4

Solution

See Also