Difference between revisions of "2006 AMC 10A Problems/Problem 7"
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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>. | 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 == | |
− | == See | + | *[[2006 AMC 12A Problems/Problem 6]] |
− | *[[2006 AMC | + | {{AMC10 box|year=2006|ab=A|num-b=6|num-a=8}} |
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Revision as of 16:25, 6 March 2007
Problem
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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 . 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 is actually half as long as the side of the square, which leads one to conclude that its value is .
See also
2006 AMC 10A (Problems • Answer Key • Resources) | ||
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 |