Problem 48 - Basic Technical Mathematics with Calculus

Question

Solve the given applied problem. The diagonal of a rectangular floor is \(3.00 \mathrm{ft}\) less than twice the length of one of the sides. If the other side is \(15.0 \mathrm{ft}\) long, what is the area of the floor?

Step-by-Step Solution

Answer

The area of the floor is 180.0 square feet.

Step 1: Define Variables

Let's define the variables for the sides of the rectangle. Let the unknown side of the rectangle be denoted as \( x \) and the diagonal as \( d \). We know the other side is \(15.0\, \mathrm{ft}\).

Step 2: Translate Statement into Equation

According to the problem, the diagonal is \(3.00\, \mathrm{ft}\) less than twice the length of one of the sides \(x\). Therefore, the equation for the diagonal is: \( d = 2x - 3.00 \).

Step 3: Apply Pythagorean Theorem

For a rectangle, the relationship involving the diagonal can be described by the Pythagorean theorem: \( d^2 = x^2 + (15.0)^2 \).

Step 4: Express Diagonal

Substitute the expression for \( d \) into the Pythagorean theorem: \((2x - 3.00)^2 = x^2 + 15.0^2 \).

Step 5: Expand and Simplify

Expand \((2x - 3.00)^2\) to \(4x^2 - 12x + 9\), and set the equation: \[ 4x^2 - 12x + 9 = x^2 + 225. \] Simplify it to: \[ 3x^2 - 12x - 216 = 0. \]

Step 6: Solve the Quadratic Equation

Solve the quadratic equation \(3x^2 - 12x - 216 = 0\). Divide through by 3: \( x^2 - 4x - 72 = 0 \). Factor the quadratic: \( (x - 12)(x + 6) = 0 \). This gives solutions \( x = 12 \) and \( x = -6 \). Since negative lengths aren't physically meaningful, \( x = 12.0\, \mathrm{ft} \).

Step 7: Calculate the Area

Now that we have both sides of the rectangle, calculate the area: \( \text{Area} = x \times 15.0 = 12.0 \times 15.0 = 180.0 \mathrm{ft}^2. \)

Key Concepts

Quadratic Equation

A quadratic equation is a powerful mathematical tool used to model various scenarios in problem-solving. It has the standard form:

• \[ ax^2 + bx + c = 0 \]

In this equation, \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. Solving these equations can help determine values of \( x \) that satisfy the equation, often representing dimensions in real-world problems.
In our exercise, we derived the quadratic equation:

• \[ 3x^2 - 12x - 216 = 0 \]

by applying the Pythagorean theorem to a rectangular floor problem. Simplifying and dividing by 3 resulted in:

• \[ x^2 - 4x - 72 = 0 \]

We found the solutions \( x = 12 \) and \( x = -6 \) through factoring. Negative lengths are non-physical, so we chose \( x = 12 \). Understanding the roots of quadratic equations is crucial to solving geometric and algebraic problems.

Rectangular Area

The area of a rectangle is a measure of the space enclosed within its sides. It is calculated with the formula:

• \[ ext{Area} = ext{length} imes ext{width} \]

For our problem, once we determined one side as \( 12.0 \) ft using the quadratic equation, and knowing the other side was \( 15.0 \) ft, we easily calculated the area.

• \[ ext{Area} = 12.0 imes 15.0 = 180.0 ext{ ft}^2 \]

Calculating the area of a rectangle is straightforward once all side lengths are known. This metric is foundational in geometry, helping us understand spatial dimensions and their applications in real life, such as flooring or lawn space.

Problem Solving

Problem solving in mathematics involves several steps to attain a solution, aligning mathematical concepts with real-world scenarios. Start by understanding the problem statement and defining variables that can represent unknowns.

• For complex problems, break them into smaller parts, like applying the Pythagorean theorem to find relationships between sides and diagonal of a rectangle.
• Convert these relationships to mathematical equations.
• Solve the equations using algebraic techniques like factoring, or applying quadratic formulas.
• Finally, evaluate the feasibility of the solutions, considering the context (e.g., dismissing negative lengths as non-sensible).

This structured approach transforms a seemingly complex question into manageable steps, ensuring clarity and accuracy in solutions. Through practice, mastering problem-solving boosts confidence in tackling various mathematical challenges.