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

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== Solution 2 ==
 
== Solution 2 ==
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The area of a 2x3 rectangle and a 3x4 rectangle combined is 18, so a 4x4 square is impossible without overlapping. Thus, the next smallest square is a 5x5, which works, so the answer is B.
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The area of a <math>2\times3</math> rectangle and a <math>3\times4</math> rectangle combined is <math>18</math>, so a <math>4\times4</math> square is impossible without overlapping. Thus, the next smallest square is a <math>5\times5</math>, which works, so the answer is <math>\boxed{B)25}</math>.
  
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Note: If you do this, always check to see if it fits, because this doesn't always work. For example, a 3x3 and a 3x4 doesn't fit into a 5x5, even though their combined area is 21.
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Note: If you do this, always check to see if it fits, because this doesn't always work. For example, a <math>3\times3</math> and a <math>3\times4</math> doesn't fit into a <math>5\times5</math>, even though their combined area is <math>21</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 22:51, 7 December 2024

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?

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

Solution 1

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 placing the two rectangles inside a $4 \times 4$ square must result in overlap, the smallest possible area of the square is $25$.

So the answer is $\boxed{\textbf{(B) }25}$.

Solution 2

The area of a $2\times3$ rectangle and a $3\times4$ rectangle combined is $18$, so a $4\times4$ square is impossible without overlapping. Thus, the next smallest square is a $5\times5$, which works, so the answer is $\boxed{B)25}$.

Note: If you do this, always check to see if it fits, because this doesn't always work. For example, a $3\times3$ and a $3\times4$ doesn't fit into a $5\times5$, even though their combined area is $21$.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

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