Problem 54 - Algebra 2 and Trigonometry

Question

The perimeter of a rectangle is 24 feet. The area of the rectangle is 32 square feet. Find the dimensions of the rectangle.

Step-by-Step Solution

Answer

The dimensions of the rectangle are 8 feet by 4 feet.

Step 1: Understand the Given Information

We are given two pieces of information: the perimeter and the area of the rectangle. The perimeter is 24 feet, and the area is 32 square feet. Let's denote the length of the rectangle as \( l \) and the width as \( w \). We have two equations to work with: the perimeter formula \( 2l + 2w = 24 \) and the area formula \( lw = 32 \).

Step 2: Set Up the Equations

Our primary goal is to solve these equations simultaneously. Start with the perimeter equation: \( 2l + 2w = 24 \). Simplify this to \( l + w = 12 \) (by dividing the entire equation by 2). Now, we have two equations: \( l + w = 12 \) (1) and \( lw = 32 \) (2).

Step 3: Express Length in Terms of Width

From equation (1), express \( l \) in terms of \( w \): \( l = 12 - w \). This expression will be useful for substituting into equation (2).

Step 4: Substitute and Solve for Width

Substitute \( l = 12 - w \) into equation (2): \( (12 - w)w = 32 \). Expand this to \( 12w - w^2 = 32 \). Rearrange to form a quadratic equation: \( w^2 - 12w + 32 = 0 \).

Step 5: Solve the Quadratic Equation

To solve the quadratic equation \( w^2 - 12w + 32 = 0 \), use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -12 \), and \( c = 32 \). Calculate the discriminant: \( b^2 - 4ac = 144 - 128 = 16 \). So, \( w = \frac{12 \pm 4}{2} \). This gives \( w = 8 \) or \( w = 4 \).

Step 6: Find Corresponding Lengths

For \( w = 8 \), substitute back into \( l = 12 - w \): \( l = 12 - 8 = 4 \). For \( w = 4 \), substitute back: \( l = 12 - 4 = 8 \).

Step 7: Verify the Solutions

Both solutions satisfy the area equation \( lw = 32 \): \( 4 \times 8 = 32 \) and \( 8 \times 4 = 32 \). Therefore, the dimensions of the rectangle are correct for both scenarios.

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle refers to the total distance around its edge. To calculate this, you add up the lengths of all its sides. Here’s the formula:

• \( P = 2l + 2w \)

where \( P \) is the perimeter, \( l \) is the length, and \( w \) is the width. In simpler terms, this formula shows us that the perimeter is twice the sum of the length and width.
For example, if the length is 5 feet and the width is 3 feet:

• \( P = 2(5) + 2(3) = 16 \) feet

This calculation highlights how the perimeter changes as we adjust the rectangle's dimensions. Understanding this concept is crucial when given a fixed perimeter, as we can explore various length-width combinations that meet this requirement.

Area of a Rectangle

The area of a rectangle is a measure of the space it occupies on a flat surface. We calculate it using the formula:

• \( A = l \times w \)

where \( A \) stands for area, \( l \) is the length, and \( w \) is the width. This formula simply multiplies the rectangle's length by its width, giving us the total square units it covers.
If you have a rectangle with a length of 4 feet and a width of 2 feet:

• \( A = 4 \times 2 = 8 \) square feet

Knowing the area helps us determine how much space a rectangle takes up, which is particularly useful in real-world applications like carpeting or tiling. When tasked with both a set area and perimeter, as in this problem, you'll work with both the area and perimeter formulas to find the rectangle's exact dimensions.

Quadratic Equations

Quadratic equations appear frequently in geometry, especially when dealing with dimensions. A quadratic equation generally takes the form:

• \( ax^2 + bx + c = 0 \)

where \( a \), \( b \), and \( c \) are constants. Solving these equations typically involves the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This solution method helps find the values of \( x \) that make the equation true.
In our exercise, once the perimeter and area equations are combined and rearranged, they result in a quadratic equation. Solving it reveals possible values for the rectangle's width (or length), which in this case are 4 feet and 8 feet. Both solutions are viable and need to be tested against the initial conditions to ensure validity.

Rectangular Dimensions

Determining rectangular dimensions involves using both perimeter and area information. With these two pieces, we can set up a system of simultaneous equations.
First, by rewriting the perimeter equation:

• \( l + w = 12 \)

we express one variable, such as the length \( l \), in terms of the other, \( w \). Then, substituting this into the area equation lets us form a quadratic equation.

• Length: \( l = 12 - w \)
• Width: Solve the quadratic formed by substitution

Through simplifying, substituting, and solving, we conclude the possible dimensions of the rectangle.
The flexibility of having two possible dimensions (8 by 4 or 4 by 8) reminds us that rectangles can often have different appearances while retaining the same area and perimeter.