2006 AMC 10A Problems/Problem 7

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Problem

2006 AMC 10A-7.GIF

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 10A-7b.GIF

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