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1978 AHSME Problems/Problem 7

Revision as of 22:07, 31 October 2020 by Justinlee2017 (talk | contribs) (added a 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 $6 \times \frac {2}{\sqrt{3}}$ $= \frac {12}{\sqrt {3}}$ $= 4\sqrt{3}$ $\boxed {E}$

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