1978 AHSME Problems/Problem 7: Difference between revisions
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 
