Difference between revisions of "2012 AMC 8 Problems/Problem 23"
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==Problem== | ==Problem== | ||
− | An equilateral triangle and a regular hexagon have equal perimeters. If the area | + | 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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==Solution 2== | ==Solution 2== | ||
− | 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 | + | 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>. |
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>. | 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>. | ||
− | + | Substitute <math>y^2\sqrt{3}</math> into the area of a regular hexagon to yield <math>\dfrac{3(4)}{2}=6</math>. | |
Therefore, our answer is <math>\boxed{\textbf{(C)}\ 6}</math>. | Therefore, our answer is <math>\boxed{\textbf{(C)}\ 6}</math>. | ||
+ | ==Solution 3== | ||
− | + | Let the side length of the triangle be <math>s</math> and the side length of the hexagon be <math>t</math>. As explained in Solution 1, <math>s=2t</math>, or <math>t=\frac{s}{2}</math>. The area of the triangle is <math>\frac{s^2\sqrt3}{4}=4</math> and the area of the hexagon is <math>\frac{t^2\sqrt3}{4} \cdot 6=\frac{3t^2\sqrt3}{2}</math>. Substituting <math>\frac{s}{2}</math> in for <math>t</math>, we get | |
− | + | <cmath>\frac{\frac{3s^2\sqrt3}{4}}{2}=\frac{3s^2\sqrt3}{8}.</cmath> | |
− | + | <math>\frac{s^2\sqrt3}{4}=4 \implies \frac{s^2\sqrt3}{8}=2 \implies \frac{3s^2\sqrt3}{8}=\boxed{\textbf{(C)}\ 6}</math>. | |
== Notes == | == Notes == | ||
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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>. | The area of a regular hexagon with side length <math>s</math> is <math>\dfrac{3s^2\sqrt{3}}{2}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/SctoIY1cbss ~savannahsolver | ||
==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}} |
Revision as of 09:37, 18 May 2022
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?
Solution 1
Let the perimeter of the equilateral triangle be . The side length of the equilateral triangle would then be and the sidelength of the hexagon would be .
A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio , 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 . The area of the hexagon is then .
Solution 2
Let the side length of the equilateral triangle be and the side length of the hexagon be . Since the perimeters are equal, we must have which reduces to . Substitute this value in to the area of an equilateral triangle to yield .
Setting this equal to gives us .
Substitute into the area of a regular hexagon to yield .
Therefore, our answer is .
Solution 3
Let the side length of the triangle be and the side length of the hexagon be . As explained in Solution 1, , or . The area of the triangle is and the area of the hexagon is . Substituting in for , we get .
Notes
The area of an equilateral triangle with side length is .
The area of a regular hexagon with side length is .
Video Solution
https://youtu.be/SctoIY1cbss ~savannahsolver
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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All AJHSME/AMC 8 Problems and Solutions |
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