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.

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 $\boxed{\textbf{(C)}}$ .

1960 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1960 AHSME Problems/Problem 4

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