Difference between revisions of "2006 AMC 10B Problems/Problem 5"
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== See Also == | == See Also == | ||
*[[2006 AMC 10B Problems]] | *[[2006 AMC 10B Problems]] | ||
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| + | *[[2006 AMC 10B Problems/Problem 4|Previous Problem]] | ||
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| + | *[[2006 AMC 10B Problems/Problem 6|Next Problem]] | ||
Revision as of 13:52, 2 August 2006
Problem
A 
rectangle and a 
rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?
Solution
By placing the rectangle adjacent to the rectangle with the 3 side of the rectangle next to the 4 side of the rectangle, we get a figure that can be completely enclosed in a square with a side length of 5. The area of this square is 
.
Since the sum of the areas of the two rectangles is 
, the area of a square cannot be less than 18. Therefore 16 is not possible.
So the answer is 
