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

## 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 AHSME (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 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 All AHSME Problems and Solutions
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