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

(New page: == Problem == Four congruent rectangles are placed as shown. The area of the outer square is <math>4</math> times that of the inner square. What is the ratio of the length of the longer si...)
 
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{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #8]] and [[2009 AMC 10A Problems|2009 AMC 10A #14]]}}
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== Problem ==
 
== Problem ==
 
Four congruent rectangles are placed as shown. The area of the outer square is <math>4</math> 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?
 
Four congruent rectangles are placed as shown. The area of the outer square is <math>4</math> 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?
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<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \sqrt {10} \qquad \textbf{(C)}\ 2 + \sqrt2 \qquad \textbf{(D)}\ 2\sqrt3 \qquad \textbf{(E)}\ 4</math>
 
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \sqrt {10} \qquad \textbf{(C)}\ 2 + \sqrt2 \qquad \textbf{(D)}\ 2\sqrt3 \qquad \textbf{(E)}\ 4</math>
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[[Category: Introductory Geometry Problems]]
  
== Solution ==
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== 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.
  
 
Then the shorter side of the rectangle is <math>1/4</math> of the side of the outer square, and the longer side of the rectangle is <math>3/4</math> of the side of the outer square, hence their ratio is <math>\boxed{3}</math>.
 
Then the shorter side of the rectangle is <math>1/4</math> of the side of the outer square, and the longer side of the rectangle is <math>3/4</math> of the side of the outer square, hence their ratio is <math>\boxed{3}</math>.
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== 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>\frac{\frac{3}{2}}{\frac{1}{2}}=\boxed{3}</math>.
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== Solution 3 ==
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Let the longer side length be <math>x</math>, and the shorter side be <math>a</math>.
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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>
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== Solution 4 ==
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WLOG, let the shorter side of the rectangle be <math>1</math>, and the long 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</math> = 4 <math>(x-2)^2</math> . Solving for x, we get the <math>x</math> = <math>3</math> .
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~coolmath2017
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2009|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2009|ab=A|num-b=7|num-a=9}}
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{{AMC10 box|year=2009|ab=A|num-b=13|num-a=15}}
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{{MAA Notice}}

Revision as of 15:02, 2 August 2020

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

$\boxed{(A)}$ 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 $2$. That too is then equivalent to $2y+1$, giving $y=1/2$. Then, the larger piece 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 long 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 the $x$ = $3$ .

~coolmath2017

See Also

2009 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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