Difference between revisions of "2012 AMC 8 Problems/Problem 23"

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<math> \textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3 </math>
 
<math> \textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3 </math>
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==See Also==
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{{AMC8 box|year=2012|num-b=22|num-a=24}}

Revision as of 10:30, 24 November 2012

An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 4, what is the area of the hexagon?

$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3$

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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
All AJHSME/AMC 8 Problems and Solutions