Difference between revisions of "2006 AMC 12A Problems/Problem 6"

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== Problem ==
 
== Problem ==
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The <math>8\times18</math> [[rectangle]] <math>ABCD</math> 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 <math>y</math>?
 
The <math>8\times18</math> [[rectangle]] <math>ABCD</math> 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 <math>y</math>?
  
 
<math> \mathrm{(A) \ } 6\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 8\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 10 </math>
 
<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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[[Image:2006 AMC 12A Problem 6.png]]
  
 
== Solution ==
 
== Solution ==

Revision as of 20:44, 15 September 2007

Problem

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 $y$?

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

2006 AMC 12A Problem 6.png

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.

Square.JPG

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

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 12 Problems and Solutions