Difference between revisions of "1978 AHSME Problems/Problem 7"
(Created page with "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>...") |
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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> | + | 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> | + | <math>\boxed {E}</math> |
Revision as of 22:07, 31 October 2020
Draw a perpendicular through the midpoint of the line of length 
such that it passes through a vertex. We now have created 
triangles. Using the ratios, we get that the hypotenuse is 
