Difference between revisions of "1956 AHSME Problems/Problem 30"
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| + | == Problem 30== | ||
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| + | If the altitude of an equilateral triangle is <math>\sqrt {6}</math>, then the area is: | ||
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| + | <math> \textbf{(A)}\ 2\sqrt{2}\qquad\textbf{(B)}\ 2\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{3}\qquad\textbf{(D)}\ 6\sqrt{2}\qquad\textbf{(E)}\ 12 </math> | ||
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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}</cmath> <cmath>= \frac{8 \sqrt{3}}{4}</cmath> <cmath>= 2 \sqrt {3}</cmath> |
<math>\boxed{B}</math> | <math>\boxed{B}</math> | ||
~JustinLee2017 | ~JustinLee2017 | ||
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| + | ==See Also== | ||
| + | {{AHSME 50p box|year=1956|num-b=29|num-a=31}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:30, 12 February 2021
Problem 30
If the altitude of an equilateral triangle is 
, then the area is:
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
See Also
| 1956 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Problem 31 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
