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.
Contents
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)
Solution1
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 
. That too is then equivalent to 
, giving 
. Then, the larger piece of the rectangles 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 | ||
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