2006 AMC 10B Problem 5

Below is the professionally curated solution for Problem 5 of the 2006 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10B solutions, or check the answer key.

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Concepts:square (geometry)rectangle optimization

Difficulty rating: 1060

A $2 \times 3$ rectangle and a $3 \times 4$ rectangle are contained within a square without overlapping at any interior 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?

$16$

$25$

$36$

$49$

$64$

Solution:

Place the rectangles side by side with their $3$ -length sides vertical. Their widths add to $2+3=5,$ and the heights $3$ and $4$ both fit within $5.$

The side cannot be smaller than $5,$ since the two smaller dimensions $2$ and $3$ must be accommodated. The smallest area is $5^2=25.$

Thus, the correct answer is B.

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