Difference between revisions of "2006 AMC 10B Problems/Problem 5"
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Since the sum of the areas of the two rectangles is <math>2\cdot3+3\cdot4=18</math>, the area of a square cannot be less than 18. Therefore 16 is not possible. | Since the sum of the areas of the two rectangles is <math>2\cdot3+3\cdot4=18</math>, the area of a square cannot be less than 18. Therefore 16 is not possible. | ||
| − | So the answer is <math>25 \Rightarrow B</math> | + | So the answer is <math>25 \Rightarrow B</math> |
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| + | Since placing the two rectangles inside a <math> 4 \times 4 </math> square must result in overlap, the smallest possible area of the square is <math>25</math>. | ||
== See Also == | == See Also == | ||
Revision as of 20:41, 2 June 2021
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 
Since placing the two rectangles inside a 
square must result in overlap, the smallest possible area of the square is 
.
See Also
| 2006 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
