2009 AMC 12A Problems/Problem 8
Problem
Four congruent rectangles are placed as shown. The area of the outer square is 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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Solution
The area of the outer square is times that of the inner square.
Therefore the side of the outer square is
times that of the inner square.
Then the shorter side of the rectangle is of the side of the outer square, and the longer side of the rectangle is
of the side of the outer square, hence their ratio is
.
See Also
2009 AMC 12A (Problems • Answer Key • 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 | |
All AMC 12 Problems and Solutions |