Problem 31 - Elementary Algebra

Question

The length of a rectangle is 2 inches more than its width. The area of the rectangle is equal to 12 inches more than three times the perimeter. Find the length and width of the rectangle.

Step-by-Step Solution

Answer

The width is 12 inches, and the length is 14 inches.

Step 1: Understand the Problem

We have a rectangle with a length that is 2 inches more than its width. We need to find the dimensions of the rectangle given that the area is 12 inches more than three times its perimeter.

Step 2: Define Variables

Let the width of the rectangle be \( w \) inches. Thus, the length of the rectangle is \( l = w + 2 \) inches.

Step 3: Set Up Expressions for Area and Perimeter

The area \( A \) of the rectangle is given by \( A = l \times w \). The perimeter \( P \) is \( P = 2(l + w) \). Substitute \( l = w + 2 \) into these equations.

Step 4: Substitute Expressions into the Given Equation

The problem states that the area is 12 inches more than three times the perimeter, so we set up the equation:\[ (w + 2)w = 12 + 3[2(w + 2) + 2w] \]

Step 5: Simplify the Equation

First, simplify the perimeter:\[ P = 2(w + 2 + w) = 2(2w + 2) = 4w + 4 \]Then, substitute it into the equation:\[ w^2 + 2w = 12 + 3(4w + 4) \]

Step 6: Solve for Width

Expand and simplify the equation:\[ w^2 + 2w = 12 + 12w + 12 \]\[ w^2 + 2w = 24 + 12w \]Rearrange to form a quadratic equation:\[ w^2 - 10w - 24 = 0 \]

Step 7: Factor the Quadratic Equation

The quadratic equation \( w^2 - 10w - 24 = 0 \) factors to:\[ (w - 12)(w + 2) = 0 \]

Step 8: Analyze the Solutions for Width

This gives us two potential solutions for \( w \):1. \( w = 12 \)2. \( w = -2 \)Since the width can't be negative, we use \( w = 12 \).

Step 9: Find Length

Using \( w = 12 \), find the length:\[ l = w + 2 = 12 + 2 = 14 \] inches.

Key Concepts

Rectangle Dimensions

When determining rectangle dimensions, start by understanding the relationship between length and width. Here, the length is 2 inches more than the width. So if the width is represented as \( w \), the length will be \( w + 2 \). This clear relationship helps to set up equations for more complex problems.

Rectangles are defined by these two measurements. This is key because once you have these two known values, you can compute other properties like perimeter and area. In this case, knowing one directly gives the other due to their relationship, which simplifies solving word problems.

Remember, keeping track of units (here, inches) for all your measurements ensures the final answer is meaningful and correct.

Quadratic Equation Solving

Quadratic equations appear in many algebra word problems, like finding rectangle dimensions by area and perimeter conditions. In our problem, we derived the quadratic equation \( w^2 - 10w - 24 = 0 \) from the relationships given. Solving these equations involves either factoring, using the quadratic formula, or completing the square.

For this example, the equation factors neatly into \( (w - 12)(w + 2) = 0 \). From factoring, each expression can equate to zero to find potential solutions for \( w \). Here are the steps:

• Factor the quadratic into two binomial expressions.
• Set each factor equal to zero: \( w - 12 = 0 \) and \( w + 2 = 0 \).
• Solve each resulting simple equation for \( w \).

The solutions, \( w = 12 \) and \( w = -2 \), need to be analyzed for practicality in the problem context. A negative width doesn't make sense, so \( w = 12 \) is the width.

Perimeter and Area Relationships

Rectangles have predictable perimeter and area relationships that help solve algebra problems. For a rectangle, the formula for the perimeter \( P \) is \( 2(l + w) \), and the area \( A \) is \( l \times w \). These properties are tied closely in problems asking for a specific condition, like our equation \( w^2 + 2w = 12 + 3(4w + 4) \).

The original task demands that the area should be 12 more than three times the perimeter. This unique statement gives a clear step toward forming a crucial equation that will lead to a quadratic form. To solve:

• Express both perimeter and area in terms of one variable \( w \).
• Set the equation so that the area matches the condition given (in this case, being more than a multiple of the perimeter).

These computations show how modifying the equations help unlock the problem's solution.