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

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==Problem==
 
==Problem==
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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?
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An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon?
  
 
<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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A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio <math> 1 : 4 </math>, since the sidelength of the small equilateral triangle is half the sidelength of the large one. Thus, the area of one of the small equilateral triangles is <math> 1 </math>. The area of the hexagon is then <math> 1 \times 6 = \boxed{\textbf{(C)}\ 6} </math>.
 
A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio <math> 1 : 4 </math>, since the sidelength of the small equilateral triangle is half the sidelength of the large one. Thus, the area of one of the small equilateral triangles is <math> 1 </math>. The area of the hexagon is then <math> 1 \times 6 = \boxed{\textbf{(C)}\ 6} </math>.
  
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==Solution 2==
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==Video Solution==
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https://youtu.be/SctoIY1cbss ~savannahsolver
  
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Let the side length of the equilateral triangle be <math>s</math> and the side length of the hexagon be <math>y</math>. Since the perimeters are equal, we must have <math>3s=6y</math> which reduces to <math>s=2y</math>. Substitute this value in to the area of an equilateral triangle to yield <math>\dfrac{(2y)^2\sqrt{3}}{4}=\dfrac{4y^2\sqrt{3}}{4}</math>.
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==Video Solution by OmegaLearn==
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https://youtu.be/j3QSD5eDpzU?t=2101
  
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Setting this equal to <math>4</math> gives us <math>\dfrac{4y^2\sqrt{3}}{4}=4\implies 4y^2\sqrt{3}=16\implies y^2\sqrt{3}=4</math>.
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~ pi_is_3.14
  
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Substitue <math>y^2\sqrt{3}</math> into the area of a regular hexagon to yield <math>\dfrac{3(4)}{2}=6</math>.
 
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Therefore, our answer is <math>\boxed{\textbf{(C)}\ 6}</math>.
 
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== Notes ==
 
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The area of an equilateral triangle with side length <math>s</math> is <math>\dfrac{s^2\sqrt{3}}{4}</math>.
 
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The area of a regular hexagon with side length <math>s</math> is <math>\dfrac{3s^2\sqrt{3}}{2}</math>.
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=22|num-a=24}}
 
{{AMC8 box|year=2012|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:21, 16 July 2024

Problem

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

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

Solution 1

Let the perimeter of the equilateral triangle be $3s$. The side length of the equilateral triangle would then be $s$ and the sidelength of the hexagon would be $\frac{s}{2}$.

A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio $1 : 4$, since the sidelength of the small equilateral triangle is half the sidelength of the large one. Thus, the area of one of the small equilateral triangles is $1$. The area of the hexagon is then $1 \times 6 = \boxed{\textbf{(C)}\ 6}$.

Video Solution

https://youtu.be/SctoIY1cbss ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=2101

~ pi_is_3.14


See Also

2012 AMC 8 (Problems • Answer Key • Resources)
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

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