Difference between revisions of "1978 AHSME Problems/Problem 7"

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

Revision as of 23:07, 31 October 2020

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 $6 \times \frac {2}{\sqrt{3}}$ $= \frac {12}{\sqrt {3}}$ $= 4\sqrt{3}$ $\boxed {E}$