Problem 41 - Algebra and Trigonometry with Analytic Geometry

Question

The perimeter of a rectangle is 40 inches, and its area is 96 in \(^{2}\). Find its length and width.

Step-by-Step Solution

Answer

The length is 12 inches and the width is 8 inches, or vice versa.

Step 1: Define the Variables

Let the length of the rectangle be \( l \) and the width be \( w \). We know the formulas for perimeter and area of a rectangle.

Step 2: Write the Perimeter Equation

The perimeter \( P \) of a rectangle is given by \( P = 2l + 2w \). From the problem, we have: \( 2l + 2w = 40 \). Dividing by 2, we get \( l + w = 20 \).

Step 3: Write the Area Equation

The area \( A \) of a rectangle is given by \( A = l \times w \). From the problem, we have: \( l \times w = 96 \).

Step 4: Solve for One Variable

From the perimeter equation, express one variable in terms of the other: \( l = 20 - w \).

Step 5: Substitute in the Area Equation

Substitute \( l = 20 - w \) into the area equation: \((20 - w)w = 96\).

Step 6: Solve the Quadratic Equation

Expand the equation: \(20w - w^2 = 96\). Rearrange to form a quadratic equation: \(w^2 - 20w + 96 = 0\).

Step 7: Find the Roots

Use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -20, c = 96 \). Solve for \( w \): \( w = \frac{20 \pm \sqrt{400 - 384}}{2} = \frac{20 \pm 4}{2} \).

Step 8: Calculate Possible Width Values

This results in solutions for \( w \): \( w = 12 \) or \( w = 8 \).

Step 9: Calculate Corresponding Lengths

For \( w = 12 \), \( l = 20 - 12 = 8 \). For \( w = 8 \), \( l = 20 - 8 = 12 \). Thus, the dimensions are interchangeable.

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle is the total distance around the outer edge of the rectangle. Imagine walking around a rectangular garden; the perimeter is the path you will follow along its edges.
To calculate the perimeter, use the formula:

• Perimeter, \( P = 2l + 2w \)

where \( l \) is the length and \( w \) is the width of the rectangle.
In our exercise, we know that the perimeter is 40 inches. Plugged into the formula, it becomes \( 2l + 2w = 40 \). We can simplify this to find \( l + w = 20 \). This equation will be helpful in determining the relationship between the length and the width later in the problem.

Area of a Rectangle

The area of a rectangle measures the space contained within its boundaries. It's as if you're interested in knowing how much paint you would need to cover a rectangular wall.
The formula to find the area is:

• Area, \( A = l \times w \)

Here, \( l \) stands for length and \( w \) for width. It's straightforward multiplication.
In the problem at hand, the area is given as 96 square inches. So we have \( l \times w = 96 \). This equation lets us express one variable in terms of the other, leveraging this in conjunction with our perimeter equation later on.

Quadratic Equations

Quadratic equations are polynomials of degree 2, typically seen in the form \( ax^2 + bx + c = 0 \). These equations can curve in a graph, creating a parabola.
To solve a quadratic equation, one popular method is the quadratic formula:

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

In our rectangle problem, we substitute from our perimeter and area equations to establish the quadratic \( w^2 - 20w + 96 = 0 \), where \( a = 1 \), \( b = -20 \), and \( c = 96 \).
When we solve this using the quadratic formula, we find possible width solutions: \( w = 12 \) and \( w = 8 \). From these, we derive the corresponding lengths, showing either is valid depending on your choice of width or length.