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−  == Problem ==
 +  #redirect [[2006 AMC 12A Problems/Problem 6]] 
−  [[Image:2006 AMC 12A Problem 6.png]]  
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−  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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−  <math> \mathrm{(A) \ } 6\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 8\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 10 </math>
 
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−  == Solution ==
 
−  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.
 
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−  [[Image:2006 AMC 12A Problem 6  Solution.png]]
 
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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>.
 
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−  == See also ==
 
−  {{AMC10 boxyear=2006ab=Anumb=6numa=8}}
 
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−  [[Category:Introductory Geometry Problems]]
 