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

(Problem)
(Solution)
Line 18: Line 18:
  
 
== Solution ==
 
== Solution ==
 
+
<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.

Revision as of 18:35, 18 November 2015

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

$\boxed{A)}$ 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 $\boxed{3}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png