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Difference between revisions of "2009 AMC 12A Problems/Problem 8"

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== Solution 1 ==
 
== Solution 1 ==
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<math>\boxed{(A)}</math>
 
 
The area of the outer square is <math>4</math> times that of the inner square.
 
The area of the outer square is <math>4</math> times that of the inner square.
 
Therefore the side of the outer square is <math>\sqrt 4 = 2</math> times that of the inner square.
 
Therefore the side of the outer square is <math>\sqrt 4 = 2</math> times that of the inner square.
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== Solution 2 ==
 
== Solution 2 ==
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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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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>. <math>2</math> is equivalent to <math>2y+1</math>, giving <math>y=1/2</math>. Then, the longer side of the rectangles is <math>3/2</math>. <math>\frac{\frac{3}{2}}{\frac{1}{2}}=\boxed{3}</math>.
  
 
== Solution 3 ==
 
== Solution 3 ==
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Hence, the answer is <math>\boxed{A}\implies 3</math>
 
Hence, the answer is <math>\boxed{A}\implies 3</math>
  
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-- Solution 4 ==
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== Solution 4 ==
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WLOG, let the shorter side of the rectangle be <math>1</math>, and the longer side be <math>x</math> . Thus, the area of the larger square is <math>(1+x)^2</math> . The area of the smaller square is <math>(x-1)^2</math> . Thus, we have the equation <math>(1+x)^2 = 4 (x-2)^2</math> . Solving for <math>x</math>, we get <math>x</math> = <math>3</math> .
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~coolmath2017
  
 
== See Also ==
 
== See Also ==

Latest revision as of 10:44, 30 October 2024

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 $4$ 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]

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \sqrt {10} \qquad \textbf{(C)}\ 2 + \sqrt2 \qquad \textbf{(D)}\ 2\sqrt3 \qquad \textbf{(E)}\ 4$

Solution 1

The area of the outer square is $4$ times that of the inner square. Therefore the side of the outer square is $\sqrt 4 = 2$ times that of the inner square.

Then the shorter side of the rectangle is $1/4$ of the side of the outer square, and the longer side of the rectangle is $3/4$ of the side of the outer square, hence their ratio is $\boxed{3}$.

Solution 2

Let the side length of the smaller square be $1$, and let the smaller side of the rectangles be $y$. Since the larger square's area is four times larger than the smaller square's, the larger square's side length is $2$. $2$ is equivalent to $2y+1$, giving $y=1/2$. Then, the longer side of the rectangles is $3/2$. $\frac{\frac{3}{2}}{\frac{1}{2}}=\boxed{3}$.

Solution 3

Let the longer side length be $x$, and the shorter side be $a$.

We have that $(x+a)^2=4(x-a)^2\implies x+a=2(x-a)\implies x+a=2x-2a\implies x=3a \implies \frac{x}{a}=3$

Hence, the answer is $\boxed{A}\implies 3$

Solution 4

WLOG, let the shorter side of the rectangle be $1$, and the longer side be $x$ . Thus, the area of the larger square is $(1+x)^2$ . The area of the smaller square is $(x-1)^2$ . Thus, we have the equation $(1+x)^2 = 4 (x-2)^2$ . Solving for $x$, we get $x$ = $3$ .

~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. AMC logo.png

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