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A 2 x3 rectangle and a 3 x4 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?

A 2 x3 rectangle and a 3 x4 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?

Grade:10

1 Answers

Arun
25763 Points
3 years ago
By experimenting, we find that the smallest side length $s$ of a square into which we can fit rectangles with dimensions $a_1\times a_2$ and $b_1\times b_2$ (with $a_1\le a_2$ and $b_1\le b_2$) is $\max(a_2,b_2,a_1+b_1)$, since one of the pairs of dimensions will have to be stacked together. In this case$s=\max(3,4,2+3)=5$,
so the area of that square will be = 52 = 25 units
Hope it helps

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