2006 AMC 12A Problems/Problem 6
- 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 1 (Shortcut)
As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.
As you can see from the diagram, the length fits into the previously blank side, so we know that it is equal to .
From there we can say so .
~Ezraft
Solution 2 (Cheap)
Because the two hexagons are congruent, we know that the perpendicular line to is half of , 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 .
~coolmath2017
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.