Difference between revisions of "2006 AMC 12A Problems/Problem 6"
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== Solution 2 (Cheap) == | == Solution 2 (Cheap) == | ||
− | Because the two hexagons are congruent, we know that the perpendicular line to A is half of BC, or <math>4</math>. Next, we plug the answer choices in to see which one works. | + | Because the two hexagons are congruent, we know that the perpendicular line to A is half of BC, or <math>4</math>. Next, we plug the answer choices in to see which one works. Trying <math>A</math>, we get the area of one hexagon is <math>72</math>, as desired, so the answer is \Longrightarrow \mathrm{(A)}$ |
== See also == | == See also == |
Revision as of 19:36, 3 May 2020
- The following problem is from both the 2006 AMC 12A #6 and 2006 AMC 10A #7, so both problems redirect to this page.
Problem
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 1
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 half the length of the side of the square, which leads to .
Solution 2 (Cheap)
Because the two hexagons are congruent, we know that the perpendicular line to A is half of BC, or . Next, we plug the answer choices in to see which one works. Trying , we get the area of one hexagon is , as desired, so the answer is \Longrightarrow \mathrm{(A)}$
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
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 (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.