# 2006 AMC 10B Problems/Problem 5

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## 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

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 $\boxed{\textbf{(B) }25}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 