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?

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

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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