Perimeter and Area | A

Question 8

On a squared grid paper (1 square = 1 square unit), make as many rectangles as you can whose lengths and widths are a whole number of units such that the area of the rectangle is 24 square units.

a. Which rectangle has the greatest perimeter?

b. Which rectangle has the least perimeter?

c. If you take a rectangle of area 32 sq cm, what will your answers be? Given any area, is it possible to predict the shape of the rectangle with the greatest perimeter as well as the least perimeter? Give examples and reasons for your answer.

Solution

Understand the Question

The area of a rectangle is given by $\text{Area} = \text{Length} \times \text{Width}$ .
For whole number dimensions, the possible lengths and widths are the factor pairs of the given area.
The perimeter is calculated using $\text{Perimeter} = 2 \times (\text{Length} + \text{Width})$ .
By comparing the perimeters of all possible rectangles for a fixed area, we can determine which dimensions give the maximum and minimum perimeters and observe general patterns.

a. Which rectangle has the greatest perimeter?

Step 1 · Find Rectangles and Perimeters for Area 24

Find all pairs of whole numbers whose product is $24$ :

$\text{Length} = 1\text{ unit}, \text{Width} = 24\text{ units}$
$\text{Length} = 2\text{ units}, \text{Width} = 12\text{ units}$
$\text{Length} = 3\text{ units}, \text{Width} = 8\text{ units}$
$\text{Length} = 4\text{ units}, \text{Width} = 6\text{ units}$

Calculate the perimeter for each using $\text{Perimeter} = 2 \times (\text{Length} + \text{Width})$ :

For $1\text{ unit} \times 24\text{ units}$ :

\begin{aligned} \text{Perimeter} &= 2 \times (1 + 24) \\[0.6em] &= 2 \times 25 \\[0.6em] &= 50\text{ units} \end{aligned}

For $2\text{ units} \times 12\text{ units}$ :

\begin{aligned} \text{Perimeter} &= 2 \times (2 + 12) \\[0.6em] &= 2 \times 14 \\[0.6em] &= 28\text{ units} \end{aligned}

For $3\text{ units} \times 8\text{ units}$ :

\begin{aligned} \text{Perimeter} &= 2 \times (3 + 8) \\[0.6em] &= 2 \times 11 \\[0.6em] &= 22\text{ units} \end{aligned}

For $4\text{ units} \times 6\text{ units}$ :

\begin{aligned} \text{Perimeter} &= 2 \times (4 + 6) \\[0.6em] &= 2 \times 10 \\[0.6em] &= 20\text{ units} \end{aligned}

Step 2 · Identify the Greatest Perimeter

Comparing the perimeters: $50\text{ units}, 28\text{ units}, 22\text{ units}, 20\text{ units}$ .

The greatest perimeter is $50\text{ units}$ , corresponding to the $1\text{ unit} \times 24\text{ units}$ rectangle.

Answer

a. The $1\text{ unit} \times 24\text{ units}$ rectangle has the greatest perimeter of $50\text{ units}$ .

b. Which rectangle has the least perimeter?

Step 1 · Identify the Least Perimeter

Comparing the perimeters: $50\text{ units}, 28\text{ units}, 22\text{ units}, 20\text{ units}$ .

The least perimeter is $20\text{ units}$ , corresponding to the $4\text{ units} \times 6\text{ units}$ rectangle.

Answer

b. The $4\text{ units} \times 6\text{ units}$ rectangle has the least perimeter of $20\text{ units}$ .

Step 1 · Find Rectangles and Perimeters for Area 32 sq cm

The factor pairs of $32$ are:

$\text{Length} = 1\text{ cm}, \text{Width} = 32\text{ cm}$
$\text{Length} = 2\text{ cm}, \text{Width} = 16\text{ cm}$
$\text{Length} = 4\text{ cm}, \text{Width} = 8\text{ cm}$

Calculate the perimeters:

For $1\text{ cm} \times 32\text{ cm}$ :

\begin{aligned} \text{Perimeter} &= 2 \times (1 + 32) \\[0.6em] &= 2 \times 33 \\[0.6em] &= 66\text{ units} \end{aligned}

For $2\text{ cm} \times 16\text{ cm}$ :

\begin{aligned} \text{Perimeter} &= 2 \times (2 + 16) \\[0.6em] &= 2 \times 18 \\[0.6em] &= 36\text{ units} \end{aligned}

For $4\text{ cm} \times 8\text{ cm}$ :

\begin{aligned} \text{Perimeter} &= 2 \times (4 + 8) \\[0.6em] &= 2 \times 12 \\[0.6em] &= 24\text{ units} \end{aligned}

Comparing perimeters ( $66, 36, 24$ ):

Greatest perimeter: $66\text{ units}$ ( $1\text{ cm} \times 32\text{ cm}$ )
Least perimeter: $24\text{ units}$ ( $4\text{ cm} \times 8\text{ cm}$ )

Step 2 · Predict the General Shape for Any Given Area

Yes, we can predict the shapes:

Greatest Perimeter: The rectangle will always be long and thin with dimensions $1 \times \text{Area}$ , because separating the factors as far apart as possible maximizes the sum $(\text{length} + \text{width})$ .
Least Perimeter: The rectangle will be closest to a square, where length and width are as close to each other as possible (or an exact square if the area is a perfect square), because minimizing the difference between dimensions minimizes the sum $(\text{length} + \text{width})$ .

Answer

c. For area $32\text{ sq cm}$ , the greatest perimeter is $66\text{ cm}$ ( $1\text{ cm} \times 32\text{ cm}$ ) and the least perimeter is $24\text{ cm}$ ( $4\text{ cm} \times 8\text{ cm}$ ). In general, the greatest perimeter is always given by a $1 \times \text{Area}$ rectangle, and the least perimeter is given by a rectangle whose dimensions are closest to a square.

Common Mistakes

Confusing Area and Perimeter: Assuming shapes with the same area must also have the same perimeter.
Missing Factor Pairs: Forgetting pairs such as $1 \times N$ when listing all possible whole-number dimensions.
Square vs. Rectangle: Forgetting that if an area is a perfect square (e.g. $36$ ), the square ( $6 \times 6$ ) provides the minimum perimeter.

Question 8

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