Difference between revisions of "2007 AMC 8 Problems/Problem 12"

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==Solution==
 
==Solution==
 
The six equilateral triangular extensions fit perfectly into the hexagon meaning the answer is <math>\boxed{\textbf{(A) }1:1}</math>
 
The six equilateral triangular extensions fit perfectly into the hexagon meaning the answer is <math>\boxed{\textbf{(A) }1:1}</math>
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 +
==Solution 2==
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Split the hexagon into six small equilateral triangles. You will see that the six outer triangles can be folded to the hexagon, so the answer is <math>\boxed{\textbf{(A) }1:1}.</math>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2007|num-b=11|num-a=13}}
 
{{AMC8 box|year=2007|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:58, 2 September 2021

Problem

A unit hexagram is composed of a regular hexagon of side length $1$ and its $6$ equilateral triangular extensions, as shown in the diagram. What is the ratio of the area of the extensions to the area of the original hexagon?

[asy] defaultpen(linewidth(0.7)); draw(polygon(3)); pair D=origin+1*dir(270), E=origin+1*dir(150), F=1*dir(30); draw(D--E--F--cycle); [/asy]

$\mathrm{(A)}\ 1:1 \qquad \mathrm{(B)}\ 6:5  \qquad \mathrm{(C)}\ 3:2 \qquad \mathrm{(D)}\ 2:1 \qquad \mathrm{(E)}\ 3:1$

Solution

The six equilateral triangular extensions fit perfectly into the hexagon meaning the answer is $\boxed{\textbf{(A) }1:1}$

Solution 2

Split the hexagon into six small equilateral triangles. You will see that the six outer triangles can be folded to the hexagon, so the answer is $\boxed{\textbf{(A) }1:1}.$

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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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