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| − | == Problem ==
| + | #redirect [[2006 AMC 12A Problems/Problem 6]] |
| − | Missing diagram
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| − | The <math>8x18</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>?
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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 ==
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| − | Since the two [[hexagon]]s 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.<br>
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| − | [[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)
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| − | == See Also ==
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| − | *[[2006 AMC 10A Problems]]
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| − | *[[2006 AMC 10A Problems/Problem 6|Previous Problem]]
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| − | *[[2006 AMC 10A Problems/Problem 8|Next Problem]]
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| − | [[Category:Introductory Geometry Problems]]
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