1978 AHSME Problems/Problem 7

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Problem 7

Opposite sides of a regular hexagon are $12$ inches apart. The length of each side, in inches, is

$\textbf{(A) }7.5\qquad \textbf{(B) }6\sqrt{2}\qquad \textbf{(C) }5\sqrt{2}\qquad \textbf{(D) }\frac{9}{2}\sqrt{3}\qquad  \textbf{(D) }\frac{9}{2}\sqrt{3}\qquad \textbf{(E) }4\sqrt{3}$


Solution

Draw a perpendicular through the midpoint of the line of length $12$ such that it passes through a vertex. We now have created $2$ $30-60-90$ triangles. Using the ratios, we get that the hypotenuse is $6 \times \frac {2}{\sqrt{3}}$ $= \frac {12}{\sqrt {3}}$ $= 4\sqrt{3} \rightarrow \boxed {\textbf{(E)}}$


See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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