Problem 53 - Algebra for College Students

Question

Set up an equation and solve each problem. The perimeter of a rectangle is 44 inches, and its area is 112 square inches. Find the length and width of the rectangle.

Step-by-Step Solution

Answer

The length is 14 inches, and the width is 8 inches (or vice versa).

Step 1: Understand the Problem

We have a rectangle with a perimeter of 44 inches and an area of 112 square inches. We need to find its length and width.

Step 2: Define Variables

Let \( l \) be the length and \( w \) be the width of the rectangle.

Step 3: Write Perimeter Equation

The perimeter of a rectangle is given by the formula \( P = 2l + 2w \). Given that the perimeter is 44 inches, we have the equation:\[ 2l + 2w = 44 \] Dividing the entire equation by 2 gives:\[ l + w = 22 \]

Step 4: Write Area Equation

The area of a rectangle is \( A = lw \). We know the area is 112 square inches, so:\[ lw = 112 \]

Step 5: Solve the System of Equations

We have two equations:1. \( l + w = 22 \)2. \( lw = 112 \)Let's solve for \( l \) from the first equation:\[ l = 22 - w \]Substitute this into the second equation:\[ (22 - w)w = 112 \]Simplify to:\[ 22w - w^2 = 112 \]Rearranging gives us a quadratic equation:\[ w^2 - 22w + 112 = 0 \]

Step 6: Solve the Quadratic Equation

Solve the quadratic equation \( w^2 - 22w + 112 = 0 \) using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -22 \), and \( c = 112 \).\[ w = \frac{22 \pm \sqrt{(-22)^2 - 4 \times 1 \times 112}}{2 \times 1} \]\[ w = \frac{22 \pm \sqrt{484 - 448}}{2} \]\[ w = \frac{22 \pm \sqrt{36}}{2} \]\[ w = \frac{22 \pm 6}{2} \]This gives: \( w = \frac{28}{2} = 14 \) or \( w = \frac{16}{2} = 8 \).

Step 7: Find Length from Width

We have two possibilities for \( w \): 14 and 8. If \( w = 14 \), then \( l = 22 - 14 = 8 \). If \( w = 8 \), then \( l = 22 - 8 = 14 \). Thus, the rectangle has length 14 inches and width 8 inches, or vice versa.

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle is the distance around the outer edge. Imagine walking along the sides of a rectangular park. You'd measure how far you walked all the way around. This distance is what's called the perimeter.

• The formula for the perimeter of a rectangle is: \( P = 2l + 2w \)
• Here, \( l \) represents the length, and \( w \) represents the width.

To find the perimeter, you simply add up all the sides. Since there are two lengths and two widths, this formula accounts for that by multiplying each by 2.
In our problem, the perimeter is given as 44 inches. Knowing this enables you to set up the equation based on the formula: \( 2l + 2w = 44 \). By simplifying, you find \( l + w = 22 \), which is more manageable for solving further.

Area of a Rectangle

The area of a rectangle is the measure of the space inside the rectangle. Think about tiling a bathroom floor: the area tells you how many tiles you need.

• The formula for the area is: \( A = lw \)
• Where \( l \) is the length and \( w \) is the width.

When you multiply the length by the width, you get the total number of square units that fit inside.
For our rectangle, knowing the area is 112 square inches allows us to express this in an equation \( lw = 112 \). This equation is crucial for finding how the length and width relate to each other, especially when you're given the perimeter as well. Together, these equations provide a system to solve.

Quadratic Equations

When you have a problem involving products of variables, such as length times width, and further complexity like a perimeter is given, you often encounter quadratic equations. Let's demystify what these are!
A quadratic equation is one where the highest power of the variable is squared, looking something like \( ax^2 + bx + c = 0 \). In our problem, to find either the length or width, we derived such an equation: \( w^2 - 22w + 112 = 0 \).

• We use the quadratic formula to solve: \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Here, \( a = 1 \), \( b = -22 \), and \( c = 112 \).

By inserting these values into the formula, you calculate possible values for \( w \). This gives us potential widths of the rectangle, from which we deduced two sets of possible dimensions for length and width: either (14, 8) or (8, 14). Solving quadratic equations can seem tough, but they are just another way of linking one variable to others.