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Difference between revisions of "2006 AMC 10B Problems/Problem 5"

 
(Added problem and solution)
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== Problem ==
 
== Problem ==
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A <math> 2 \times 3 </math> rectangle and a <math> 3 \times 4 </math> 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?
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<math> \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 25\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 49\qquad \mathrm{(E) \ } 64 </math>
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== Solution ==
 
== Solution ==
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By placing the <math> 2 \times 3 </math> rectangle adjacent to the <math> 3 \times 4 </math> rectangle with the 3 side of the <math> 2 \times 3 </math> rectangle next to the 4 side of the <math> 3 \times 4 </math> 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 <math>5^2 = 25</math>
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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.
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So the answer is <math>25 \Rightarrow E</math>
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== See Also ==
 
== See Also ==
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]

Revision as of 22:23, 13 July 2006

Problem

A $2 \times 3$ rectangle and a $3 \times 4$ 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?

$\mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 25\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 49\qquad \mathrm{(E) \ } 64$

Solution

By placing the $2 \times 3$ rectangle adjacent to the $3 \times 4$ rectangle with the 3 side of the $2 \times 3$ rectangle next to the 4 side of the $3 \times 4$ 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 $5^2 = 25$

Since the sum of the areas of the two rectangles is $2\cdot3+3\cdot4=18$ the area of a square cannot be less than 18. Therefore 16 is not possible.

So the answer is $25 \Rightarrow E$

See Also

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