Difference between revisions of "2009 AMC 12A Problems/Problem 8"
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Revision as of 11:28, 2 July 2009
- 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?
![[asy] unitsize(6mm); defaultpen(linewidth(.8pt)); path p=(1,1)--(-2,1)--(-2,2)--(1,2); draw(p); draw(rotate(90)*p); draw(rotate(180)*p); draw(rotate(270)*p); [/asy]](http://latex.artofproblemsolving.com/4/f/8/4f8d08abd119d663ad6630516ebc1b3bd784727e.png)
Solution
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 
.
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 | ||
