Difference between revisions of "2006 AMC 10A Problems/Problem 7"
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== Solution == | == 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*8=144. This means the square will have four sides of length 12. The only way to do this is shown below. | 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*8=144. 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:Square.JPG]] | [[Image:Square.JPG]] | ||
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| + | As you can see from the diagram, the line segment denoted as ''y'' is actually one half the length of the square, which leads one to conclude that its value is 12/2 = 6, (A) | ||
== See Also == | == See Also == | ||
*[[2006 AMC 10A Problems]] | *[[2006 AMC 10A Problems]] | ||
Revision as of 13:30, 18 July 2006
Problem
Missing diagram
The 
rectangle 
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 
?
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*8=144. This means the square will have four sides of length 12. The only way to do this is shown below.
As you can see from the diagram, the line segment denoted as y is actually one half the length of the square, which leads one to conclude that its value is 12/2 = 6, (A)
