2009 AMC 12A Problems/Problem 8
- The following problem is from both the 2009 AMC 12A #8 and 2009 AMC 10A #14, so both problems redirect to this page.
Problem
Four congruent rectangles are placed as shown. The area of the outer square is times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?
Solution 1
The area of the outer square is times that of the inner square. Therefore the side of the outer square is times that of the inner square.
Then the shorter side of the rectangle is of the side of the outer square, and the longer side of the rectangle is of the side of the outer square, hence their ratio is .
Solution 2
Let the side length of the smaller square be , and let the smaller side of the rectangles be . Since the larger square's area is four times larger than the smaller square's, the larger square's side length is . is equivalent to , giving . Then, the longer side of the rectangles is . .
Solution 3
Let the longer side length be , and the shorter side be .
We have that
Hence, the answer is
Solution 4
WLOG, let the shorter side of the rectangle be , and the longer side be . Thus, the area of the larger square is . The area of the smaller square is . Thus, we have the equation . Solving for , we get = .
~coolmath2017
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 |
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.