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Difference between revisions of "1960 AHSME Problems/Problem 4"

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==Solution==
 
==Solution==
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If two of the angles are <math>60^{\circ}</math>, then the other angle is <math>60^{\circ}</math> because angles in triangle add up to <math>180^{\circ}</math>.  That makes the triangle an equilateral triangle, so all sides are <math>4</math> inches long.
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If two of the angles are <math>60^{\circ}</math>, then the other angle is <math>60^{\circ}</math> because angles in triangle add up to <math>180^{\circ}</math>.  That makes the triangle an [[equilateral triangle]], so all sides are <math>4</math> inches long.
  
 
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==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1960|num-b=3|num-a=5}}
 
{{AHSME 40p box|year=1960|num-b=3|num-a=5}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 17:54, 17 May 2018

Problem

Each of two angles of a triangle is $60^{\circ}$ and the included side is $4$ inches. The area of the triangle, in square inches, is:

$\textbf{(A)} 8\sqrt{3}\qquad \textbf{(B)} 8\qquad \textbf{(C)} 4\sqrt{3}\qquad \textbf{(D)} 4\qquad \textbf{(E)} 2\sqrt{3}$

Solution

If two of the angles are $60^{\circ}$, then the other angle is $60^{\circ}$ because angles in triangle add up to $180^{\circ}$. That makes the triangle an equilateral triangle, so all sides are $4$ inches long.

[asy] draw((0,0)--(50,0)--(25,43.301)--cycle); label("$4$",(10,25)); label("$2$",(12.5,-5)); label("$2$",(37.5,-5)); label("$4$",(40,25)); draw((25,43.301)--(25,0)); label("$2\sqrt{3}$",(20,15)); draw((25,3)--(28,3)--(28,0)); [/asy]

Using the area formula $A = \frac{s^2\sqrt{3}}{4}$, the area of the triangle is $\frac{4^2\sqrt{3}}{4} = 4\sqrt{3}$ square inches, which is answer choice $\boxed{\textbf{(C)}}$.

See Also

1960 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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All AHSME Problems and Solutions
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