2006 AMC 10B Problems/Problem 5

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 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.

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.

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 10 Problems and Solutions

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2006 AMC 10B Problems/Problem 5

Contents

Problem

Solution 1

Solution 2

See Also