To solve this problem, we'll use the information that Casey has 24 feet of fencing. The perimeter of a rectangle is given by the formula:

$P = 2L + 2W$

where $L$ is the length and $W$ is the width. Substituting $P = 24$:

$2L + 2W = 24$

Simplifying, we find:

$L + W = 12$

This equation shows that the sum of the length and the width must be 12 feet. By assigning different values to $L$ and $W$ while ensuring their sum equals 12, we can construct different rectangles. For each rectangle, the area is calculated as:

$A = L \times W$

Rectangles

1. Rectangle 1

   • Let $L = 8$ feet and $W = 4$ feet.
   • Check: $L + W = 8 + 4 = 12$.
   • Area: $A = 8 \times 4 = 32$ square feet.

2. Rectangle 2

   • Let $L = 6$ feet and $W = 6$ feet.
   • Check: $L + W = 6 + 6 = 12$.
   • Area: $A = 6 \times 6 = 36$ square feet.

Visualization

I'll describe the two rectangles so you can draw them:

• Rectangle 1: Label one side as 8 feet and the adjacent side as 4 feet.
• Rectangle 2: Label both sides as 6 feet (a square).

If you'd like, I can create an image illustrating these rectangles.

Recap of Areas

1. Rectangle 1: 32 square feet.
2. Rectangle 2: 36 square feet.

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Would you like a visual representation of these rectangles?

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Related Questions

1. What happens to the area if we make one side extremely short?
2. Can a rectangle with $L = 11$ feet and $W = 1$ foot be constructed with the given fencing?
3. How does changing the dimensions affect the area of the rectangle?
4. What would the dimensions be if we wanted the smallest area possible?
5. Could Casey make a circular fence with 24 feet of fencing?

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Tip: To maximize the area of a rectangle with a fixed perimeter, aim for a square!