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

(Solution to 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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<asy>
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draw((0,0)--(50,0)--(25,43.301)--cycle);
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label("$4$",(10,25));
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label("$2$",(12.5,-5));
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label("$2$",(37.5,-5));
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label("$4$",(40,25));
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draw((25,43.301)--(25,0));
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label("$2\sqrt{3}$",(20,15));
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draw((25,3)--(28,3)--(28,0));
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</asy>
  
 
Using the area formula <math>A = \frac{s^2\sqrt{3}}{4}</math>, the area of the triangle is <math>\frac{4^2\sqrt{3}}{4} = 4\sqrt{3}</math> square inches, which is answer choice <math>\boxed{\textbf{(C)}}</math>.
 
Using the area formula <math>A = \frac{s^2\sqrt{3}}{4}</math>, the area of the triangle is <math>\frac{4^2\sqrt{3}}{4} = 4\sqrt{3}</math> square inches, which is answer choice <math>\boxed{\textbf{(C)}}</math>.
  
 
==See Also==
 
==See Also==
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{{AHSME box|year=1960|num-b=3|num-a=5}}
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{{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
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
All AHSME Problems and Solutions
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