2009 AMC 12A Problems/Problem 8

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(Redirected from 2009 AMC 10A Problems/Problem 14)

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?

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);...

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

Solution

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 $\boxed{3}$ .

2009 AMC 12A (Problems • 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

2009 AMC 10A (Problems • 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

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2009 AMC 12A Problems/Problem 8

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