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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
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~JustinLee2017
 
~JustinLee2017
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==See Also==
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{{AHSME 50p box|year=1956|num-b=29|num-a=31}}
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{{MAA Notice}}

Latest revision as of 22:30, 12 February 2021

Problem 30

If the altitude of an equilateral triangle is $\sqrt {6}$, then the area is:

$\textbf{(A)}\ 2\sqrt{2}\qquad\textbf{(B)}\ 2\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{3}\qquad\textbf{(D)}\ 6\sqrt{2}\qquad\textbf{(E)}\ 12$

Solution

Drawing the altitude, we see that is opposite the $60^{\circ}$ angle in a $30-60-90$ triangle. Thus, we find that the shortest leg of that triangle is $\sqrt{\frac{6}{3}} = \sqrt{2}$, and the side of the equilateral triangle is then $2\sqrt{2}$. Thus, the area is \[\frac{(2\sqrt 2)^2 \cdot \sqrt{3}}{4}\] \[= \frac{8 \sqrt{3}}{4}\] \[= 2 \sqrt {3}\] $\boxed{B}$

~JustinLee2017


See Also

1956 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 31
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All AHSME Problems and Solutions

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