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> | ||
− | ==Solution== | + | ==Solution 1== |
Let the perimeter of the equilateral triangle be <math> 3s </math>. The side length of the equilateral triangle would then be <math> s </math> and the sidelength of the hexagon would be <math> \frac{s}{2} </math>. | Let the perimeter of the equilateral triangle be <math> 3s </math>. The side length of the equilateral triangle would then be <math> s </math> and the sidelength of the hexagon would be <math> \frac{s}{2} </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>. | 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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+ | 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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+ | 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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+ | 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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+ | ---- | ||
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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}} |
Revision as of 20:38, 9 November 2013
Contents
Problem
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?
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 $\dfrac{(2y)^2\sqrt{3}}{4}=\dfrac{4y^2\sqrt{3}}{4}}$ (Error compiling LaTeX. Unknown error_msg).
Setting this equal to gives us .
Substitue into the area of a regular hexagon to yield .
Therefore, our answer is .
Notes
The area of an equilateral triangle with side length is .
The area of a regular hexagon with side length is .
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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