2010 CEMC Gauss (Grade 8) Problems/Problem 5

Problem

The area of a rectangle is $12$ . Each of its side lengths is a whole number. What is the smallest possible perimeter of this rectangle?

$\text{ (A) }\ 24 \qquad\text{ (B) }\ 48 \qquad\text{ (C) }\ 26 \qquad\text{ (D) }\ 14 \qquad\text{ (E) }\ 16$

Solution 1

The product of the side lengths of the rectangle is its area. Because we know both are integers, we can look at the factor pairs of $12$ . The numbers in the pairs will represent the side lengths of each rectangle, and we can find the area.

The factor pairs of $12$ are $(1, 12)$ , $(2, 6)$ , and $(3, 4)$ .

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Let $p$ be the perimeter of each rectangle. For $(1, 12)$ , the perimeter is then:

$p = 2 \times 1 + 2 \times 12 = 2 + 2 \times 12 = 2 + 24 = 26$

For $(2, 6)$ , the perimeter is:

$p = 2 \times 2 + 2 \times 6 = 4 + 2 \times 6 = 4 + 12 = 16$

For $(3, 4)$ , the perimeter is:

$p = 2 \times 3 + 2 \times 4 = 6 + 2 \times 4 = 6 + 8 = \boxed {\textbf {(D) } 14}$

~anabel.disher

Solution 1.5

We can notice that the rectangle with the smallest perimeter will be the one whose side length sum is the smallest. We can show this through the formula for the perimeter of the rectangle:

$p = 2l + 2w = 2(l + w)$

To minimize the perimeter, $l + w$ will have to be minimized. Using the same process with the factor pairs, we can see that $(3, 4)$ has the lowest sum. This means that we can simply find the perimeter of that rectangle without needing to calculate the perimeter of the other ones.

$p = 2(l + w) = 2(3 + 4) = 2 \times 7 = \boxed {\textbf {(D) } 14}$

~anabel.disher

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2010 CEMC Gauss (Grade 8) Problems/Problem 5

Problem

Solution 1

Solution 1.5