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}</cmath> <cmath>= \frac{8 \sqrt{3}}{4}</cmath> <cmath>= 2 \sqrt {3 | + | <cmath>\frac{(2\sqrt 2)^2 \cdot \sqrt{3}}{4}</cmath> <cmath>= \frac{8 \sqrt{3}}{4}</cmath> <cmath>= 2 \sqrt {3}</cmath> |
<math>\boxed{B}</math> | <math>\boxed{B}</math> | ||
~JustinLee2017 | ~JustinLee2017 | ||
Revision as of 18:01, 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
![\[\frac{(2\sqrt 2)^2 \cdot \sqrt{3}}{4}\]](http://latex.artofproblemsolving.com/0/9/b/09b8976582aac10c542aa289ba0c10fe5cf6abf1.png)
![\[= \frac{8 \sqrt{3}}{4}\]](http://latex.artofproblemsolving.com/c/5/2/c52b0c651d48bdb0308ac4e4fece9ecb24336a87.png)
![\[= 2 \sqrt {3}\]](http://latex.artofproblemsolving.com/1/7/2/17245af2e53c0f55d07519f00cc4610e0220803d.png)
~JustinLee2017
