Difference between revisions of "2006 AMC 10A Problems/Problem 7"

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== Problem ==
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#redirect [[2006 AMC 12A Problems/Problem 6]]
{{image}}
 
 
 
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>
 
 
 
== 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.<br>
 
 
 
[[Image:Square.JPG]]
 
 
 
As you can see from the diagram, the [[line segment]] denoted as <math>y</math> is actually half as long as the side of the square, which leads one to conclude that its value is <math>\frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}</math>.
 
 
 
 
 
== See Also ==
 
*[[2006 AMC 10A Problems]]
 
 
 
*[[2006 AMC 10A Problems/Problem 6|Previous Problem]]
 
 
 
*[[2006 AMC 10A Problems/Problem 8|Next Problem]]
 
 
 
[[Category:Introductory Geometry Problems]]
 

Latest revision as of 22:38, 27 April 2008

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