2006 AMC 10B Problem 5

Problem 5 of 25EasierAlgebra Geometry

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?

Show solution

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.

Because the rectangles are axis-aligned and their interiors do not overlap, their horizontal projections or their vertical projections must be disjoint. In either direction, the first rectangle spans at least

2

and the second spans at least

3,

so the square’s side is at least

2+3=5.

The smallest area is therefore

5^2=25.

Thus, the correct answer is B.

Answer choices

Solution

More practice