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

(Added Image #2 and small changes)
(Duplicate, AMC 10 box)
Line 1: Line 1:
 +
{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2008 AMC 10A #7]]}}
 
== Problem ==
 
== Problem ==
 
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>
  
 
[[Image:2006 AMC 12A Problem 6.png]]
 
[[Image:2006 AMC 12A Problem 6.png]]
Line 14: Line 15:
  
 
== See also ==
 
== See also ==
*[[2006 AMC 10A Problems/Problem 7]]
 
 
{{AMC12 box|year=2006|ab=A|num-b=5|num-a=7}}
 
{{AMC12 box|year=2006|ab=A|num-b=5|num-a=7}}
 +
{{AMC10 box|year=2006|ab=A|num-b=6|num-a=8}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 22:40, 27 April 2008

The following problem is from both the 2006 AMC 12A #6 and 2008 AMC 10A #7, so both problems redirect to this page.

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.

2006 AMC 12A Problem 6 - Solution.png

As you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y$$= \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
2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 10 Problems and Solutions