Problem 83 - Elementary Algebra

Question

The width of a rectangle is \(6 \mathrm{ft}\) less than its length. Its area is \(112 \mathrm{ft}^{2}\).

Step-by-Step Solution

Answer

The length is 14 ft and the width is 8 ft.

Step 1: Define variables

Let the length of the rectangle be denoted as \( l \) and the width as \( w \). According to the problem, the width is 6 feet less than the length. Therefore, \( w = l - 6 \).

Step 2: Write the area formula

The area of a rectangle is given by the formula \( \text{Area} = l \times w \). We are given that the area is 112 square feet. Thus, \( l \times w = 112 \).

Step 3: Substitute the width

Using the expression for the width from Step 1, substitute \( w = l - 6 \) into the area equation: \( l \times (l - 6) = 112 \).

Step 4: Set up the equation

Expand the equation to obtain a quadratic equation: \[ l^2 - 6l = 112 \]

Step 5: Simplify the quadratic equation

Move all terms to one side to set the equation to zero: \[ l^2 - 6l - 112 = 0 \]

Step 6: Solve the quadratic equation

Solve the quadratic equation using the quadratic formula \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -6 \), and \( c = -112 \). Substitute these values into the formula: \[ l = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times (-112)}}{2 \times 1} \] Simplify inside the square root: \[ l = \frac{6 \pm \sqrt{36 + 448}}{2} \] \[ l = \frac{6 \pm \sqrt{484}}{2} \] Since \( \text{sqrt}(484) = 22 \): \[ l = \frac{6 \pm 22}{2} \] This gives two solutions: \[ l = \frac{6 + 22}{2} = 14 \] \[ l = \frac{6 - 22}{2} = -8 \] Since a length cannot be negative, the valid solution is \( l = 14 \).

Step 7: Find the width

Substitute \( l = 14 \) back into the width equation from Step 1: \( w = l - 6 = 14 - 6 = 8 \).

Key Concepts

area of a rectangle

The area of a rectangle is a fundamental concept in geometry. To find the area, you simply multiply the length by the width. The formula is expressed as \(\text{Area} = l \times w\).
For example, if a rectangle has a length of 10 feet and a width of 5 feet, the area would be \(\text{Area} = 10 \times 5 = 50 \text{ square feet}\). This concept is easy to understand and is widely applicable to different problems in mathematics.

quadratic formula

The quadratic formula is a powerful tool for solving equations of the form \(ax^2 + bx + c = 0\). The formula is expressed as: \[-b \pm \sqrt{b^2 - 4ac} / 2a \].
Using this formula involves a few straightforward steps:

• Identify coefficients \(a, b, \) and \(c\).
• Substitute them into the formula.
• Simplify to find the solutions.

substitution in algebra

Substitution is an essential method in algebra for solving equations. It involves replacing a variable with an equivalent expression.
In the given problem, we knew that the width \(w\) was 6 feet less than the length \(l\). So, we substituted \(w = l - 6\) in the area formula \(l \times w = 112 \text{ square feet} \). This helps convert a complex problem into a simpler one that we can solve more easily.

variable definition

Defining variables is a crucial step in solving mathematical problems as it helps clarify what each symbol represents.
For instance, in the problem, we defined \(l\) as the length of the rectangle and \(w\) as the width. It’s also given that \(w = l - 6\). By defining variables and relationships correctly, the subsequent steps become more manageable and accurate.