Difference between revisions of "1956 AHSME Problems/Problem 30"
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==Solution== | ==Solution== | ||
Drawing the altitude, we see that is opposite the <math>60^{\circ}</math> angle in a <math>30-60-90</math> triangle. Thus, we find that the shortest leg of that triangle is <math>\sqrt{\frac{6}{3}} = \sqrt{2}</math>, and the side of the equilateral triangle is then <math>2\sqrt{2}</math>. Thus, the area is | Drawing the altitude, we see that is opposite the <math>60^{\circ}</math> angle in a <math>30-60-90</math> triangle. Thus, we find that the shortest leg of that triangle is <math>\sqrt{\frac{6}{3}} = \sqrt{2}</math>, and the side of the equilateral triangle is then <math>2\sqrt{2}</math>. Thus, the area is | ||
| − | <cmath>\frac{ (2\sqrt 2)^2 \cdot \sqrt{3}}{4} = \frac{8 \sqrt 3}{4} = 2 \sqrt 3}</cmath> | + | <cmath>\frac{(2\sqrt 2)^2 \cdot \sqrt{3}}{4} = \frac{8 \sqrt{3}}{4} = 2 \sqrt {3}}</cmath> |
<math>\boxed{B}</math> | <math>\boxed{B}</math> | ||
~JustinLee2017 | ~JustinLee2017 | ||
Revision as of 18:00, 12 February 2021
Solution
Drawing the altitude, we see that is opposite the 
angle in a 
triangle. Thus, we find that the shortest leg of that triangle is 
, and the side of the equilateral triangle is then 
. Thus, the area is
\[\frac{(2\sqrt 2)^2 \cdot \sqrt{3}}{4} = \frac{8 \sqrt{3}}{4} = 2 \sqrt {3}}\] (Error compiling LaTeX. Unknown error_msg)
~JustinLee2017
