Difference between revisions of "2009 AMC 12A Problems/Problem 8"
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== Solution 2 == | == Solution 2 == | ||
Let the side length of the smaller square be <math>1</math>, and let the smaller side of the rectangles be <math>y</math>. Since the larger square's area is four times larger than the smaller square's, the larger square's side length is <math>2</math>. That too is then equivalent to <math>2y+1</math>, giving <math>y=1/2</math>. Then, the larger piece of the rectangles is <math>3/2</math>. <math>3/2/1/2=\boxed{3}</math>. | Let the side length of the smaller square be <math>1</math>, and let the smaller side of the rectangles be <math>y</math>. Since the larger square's area is four times larger than the smaller square's, the larger square's side length is <math>2</math>. That too is then equivalent to <math>2y+1</math>, giving <math>y=1/2</math>. Then, the larger piece of the rectangles is <math>3/2</math>. <math>3/2/1/2=\boxed{3}</math>. | ||
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+ | == Solution 3 == | ||
+ | Let the longer side length be <math>x</math>, and the shorter side be <math>a</math>. | ||
+ | |||
+ | We have that <math>(x+a)^2=4(x-a)^2\implies x+a=2(x-a)\implies x+a=2x-2a\implies x=3a \implies \frac{x}{a}=3</math> | ||
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+ | Hence, the answer is <math>\boxed{A}\implies 3</math> | ||
== See Also == | == See Also == |
Revision as of 17:43, 15 July 2017
- The following problem is from both the 2009 AMC 12A #8 and 2009 AMC 10A #14, so both problems redirect to this page.
Contents
[hide]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?
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 . .
Solution 3
Let the longer side length be , and the shorter side be .
We have that
Hence, the answer 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.