Problem 52 - Algebra 2 and Trigonometry

Question

The area of a rectangular rug is 48 square feet. The length of the rug is 2 feet longer than the width. What are the dimensions of the rug?

Step-by-Step Solution

Answer

The rug’s dimensions are 6 feet by 8 feet.

Step 1: Understanding the Problem

We need to find the dimensions of a rectangular rug, given that the area is 48 square feet and the length is 2 feet longer than the width. Let's denote the width as \( w \) and the length as \( l = w + 2 \).

Step 2: Setting Up the Equation

The area of the rectangle is given by the formula \( ext{Area} = ext{length} \times ext{width} \). Therefore, \( 48 = (w+2) \times w \).

Step 3: Solving the Quadratic Equation

We expand the equation: \( (w+2)w = 48 \) becomes \( w^2 + 2w = 48 \). Rearranging gives us \( w^2 + 2w - 48 = 0 \).

Step 4: Finding the Roots of the Equation

We can use the quadratic formula, \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), and \( c = -48 \). This will help us find \( w \).

Step 5: Calculating the Discriminant

Compute the discriminant: \( b^2 - 4ac = 2^2 - 4(1)(-48) = 4 + 192 = 196 \).

Step 6: Applying the Quadratic Formula

Plug the values into the quadratic formula: \( w = \frac{-2 \pm \sqrt{196}}{2} \). Therefore, \( w = \frac{-2 \pm 14}{2} \).

Step 7: Determining the Width

We have two possible solutions for \( w \): \( \frac{12}{2} = 6 \) and \( \frac{-16}{2} = -8 \). Since \( w \) cannot be negative, \( w = 6 \) feet.

Step 8: Finding the Length

Using \( l = w + 2 \), we find \( l = 6 + 2 = 8 \) feet.

Key Concepts

Area of Rectangles

When it comes to geometry, understanding the area of rectangles is essential. A rectangle is a four-sided shape with opposite sides that are equal in length. The formula to find the area is pretty straightforward:
\[\text{Area} = \text{Length} \times \text{Width}.\]

• **Length** is the longer side of the rectangle.
• **Width** is the shorter side of the rectangle.
• The **area** is the space contained within the perimeters.

Let's apply this to a practical example. Consider a rug. If you know the area and one dimension, you can determine the other. For instance, if a rug's area is 48 square feet and the length is 2 feet longer than the width, you can set up a simple equation:
Let the width be \( w \), then the length will be \( w+2 \). The equation becomes \( 48 = w \times (w+2) \). This equation will help us figure out the dimensions.

Problem Solving in Mathematics

Problem-solving is at the heart of mathematics. It involves understanding the problem, devising an effective strategy, carrying out the calculations, and checking the results. Let's break it down:

• **Understand the Problem**: Determine what is being asked. For the rug question, it was the rug's dimensions.
• **Set Up the Equation**: Translate the word problem into a mathematical expression. Here, know that area = length × width gives us \( 48 = w(w+2) \).
• **Solve the Equation**: Simplification often leads to a quadratic equation, such as \( w^2 + 2w - 48 = 0 \).
• **Verify**: Once you have a solution, plug the numbers back into the context of the problem to ensure accuracy. Ensuring \( w = 6 \) works because the length \( l = 8 \) yields an area of 48 sq ft.

The satisfaction comes from solving the puzzle, validating your logical approach, and expanding your problem-solving toolkit.

Discriminant in Quadratic Formula

The discriminant in the quadratic formula might sound complex, but it's quite helpful. It helps us find the nature of roots without solving the entire equation. For any quadratic equation \( ax^2 + bx + c = 0 \), the discriminant is given by:
\[D = b^2 - 4ac.\]Upon calculation,

• If \( D > 0 \), two distinct real roots exist.
• If \( D = 0 \), there's exactly one real root (a perfect square).
• If \( D < 0 \), no real roots exist, only complex ones.

For the rug example, the discriminant was computed as 196.
Since \( D = 196 > 0 \), we expected two real roots. Calculating gives \( w = 6 \) or \( w = -8 \). Only \( w = 6 \) is valid for real-world problems as measurements cannot be negative. This illustrates how the discriminant simplifies our understanding of solutions.