# Difference between revisions of "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.

## 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 $2$. $2$ 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

 2009 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 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 byProblem 13 Followed byProblem 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