Difference between revisions of "2006 AMC 10A Problems/Problem 7"

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== Problem ==
 
== Problem ==
[[Image:2006_AMC_10A-7.GIF]]
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[[Image:2006 AMC 12A Problem 6.png]]
  
 
The <math>8\times18</math> [[rectangle]] <math>ABCD</math> is cut into two [[congruent]] [[hexagon]]s, as shown, in such a way that the two hexagons can be repositioned without overlap to form a [[square (geometry)|square]].  What is <math>y</math>?
 
The <math>8\times18</math> [[rectangle]] <math>ABCD</math> is cut into two [[congruent]] [[hexagon]]s, as shown, in such a way that the two hexagons can be repositioned without overlap to form a [[square (geometry)|square]].  What is <math>y</math>?
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Since the two [[hexagon]]s are going to be repositioned to form a [[square (geometry) | square]] without overlap, the [[area]] will remain the same. The rectangle's area is <math>18\cdot8=144</math>. This means the square will have four sides of length 12. The only way to do this is shown below.
 
Since the two [[hexagon]]s are going to be repositioned to form a [[square (geometry) | square]] without overlap, the [[area]] will remain the same. The rectangle's area is <math>18\cdot8=144</math>. This means the square will have four sides of length 12. The only way to do this is shown below.
  
[[Image:2006_AMC_10A-7b.GIF]]
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[[Image:2006 AMC 12A Problem 6 - Solution.png]]
  
As you can see from the diagram, the [[line segment]] denoted as <math>y</math> is actually half as long as the side of the square, which leads one to conclude that its value is <math>\frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}</math>.
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As you can see from the diagram, the [[line segment]] denoted as <math>y</math> is half the length of the side of the square, which leads to <math>y</math><math> = \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 20:02, 15 September 2007

Problem

2006 AMC 12A Problem 6.png

The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$?

$\mathrm{(A) \ } 6\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 8\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 10$

Solution

Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$. This means the square will have four sides of length 12. The only way to do this is shown below.

2006 AMC 12A Problem 6 - Solution.png

As you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y$$= \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}$.

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 AMC 10 Problems and Solutions
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