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

m (Solution)
m (See Also)
Line 22: Line 22:
  
 
==See Also==
 
==See Also==
{{AHSME box|year=1960|num-b=3|num-a=5}}
+
{{AHSME 40p box|year=1960|num-b=3|num-a=5}}

Revision as of 20:11, 10 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 (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS