2006 AMC 10B Problem 5

Lower bound: the rectangles have areas $6$ and $12$, so the square must have area at least $18$. A $4 \times 4$ square has area $16 < 18$, so side $4$ is impossible, and any smaller side is too.

Construction for side $5$: stand the $3 \times 4$ rectangle with its $3$-side horizontal and $4$-side vertical, and stand the $2 \times 3$ rectangle next to it with its $2$-side horizontal and $3$-side vertical. Together they span width $3 + 2 = 5$ and height $\max(4, 3) = 4$, which fits inside a $5 \times 5$ square.

So the smallest square has area $5^2 = 25$.

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

A minimum-size question has two halves: a bound that shows nothing smaller works, and an explicit arrangement showing the bound is achievable. Area gives the quickest bound; for the construction, orient the rectangles so their long sides run the same direction and their short sides add up.

• Stacking the rectangles end to end (3 + 4 = 7) and answering 49, without trying the side-by-side arrangement.
• Concluding that area $18 > 16$ alone proves $25$, without confirming the rectangles actually fit in a $5 \times 5$ square (a $3\times3$ and $3\times4$ pair, for instance, would not).