Problem 59 - Introductory Algebra for College Students

Question

The length of a rectangle is 3 meters longer than the width. If the area is 36 square meters, find the rectangle's dimensions. Round to the nearest tenth of a meter.

Step-by-Step Solution

Answer

The dimensions of the rectangle are \(w \approx\) 4.6 meters for the width and \(w + 3 \approx\) 7.6 meters for the length.

Step 1: Define Variables

Let the width of the rectangle be \(w\) meters. Then the length of the rectangle is \(w+3\) meters according to the problem statement.

Step 2: Set up Equation

Next, set up the equation using the formula for the area of a rectangle which is Area = Length * Width. As given, the area of the rectangle is 36 sq. meters. So, \( w * (w + 3) = 36 \).

Step 3: Solve the Equation

Solving this quadratic equation, we get \( w^2 + 3w - 36 = 0 \). Using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we get two possible solutions for width. However, width can't be negative, so we discard the negative solution and select only the positive value.

Step 4: Calculate the Length

Plug the width found in Step 3 into the expression for the length to calculate the length of the rectangle.

Key Concepts

Rectangular Dimensions

When tackling problems involving rectangles, understanding the relationship between the length and the width is crucial. In this exercise, we started by defining the width of the rectangle as \(w\). Given that the length is 3 meters more than the width, the length can be represented as \(w + 3\). This transformation helps simplify our understanding of the rectangle's dimensions:

• Length = Width + 3 meters
• Width = \(w\)

These expressions allow us to write equations that define the problem in mathematical terms. By representing dimensions in a formulaic way, it becomes easier to solve complex algebraic problems using elementary operations. Knowing the exact relationship between the length and width is the foundation for formulating and solving the equations representing geometric scenarios.

Algebraic Problem-Solving

Algebraic problem-solving in this scenario involves setting up an equation based on real-world conditions and then solving it using algebraic techniques and formulas. Initially, we formed the equation \(w \cdot (w + 3) = 36\), reflecting the given area of the rectangle. This step translates a verbal description of the problem into mathematical language. Once we have the equation:

• Expand and rearrange: \(w^2 + 3w - 36 = 0\)
• Apply the quadratic formula \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Solving yields one potential negative and one positive solution for \(w\)

Only the positive solution is considered valid for the width, reinforcing the importance of interpreting results in a logical, applicable manner. This ensures that the dimensions make sense in the context of the problem.

Area Calculation

Calculating the area of a rectangle is a fundamental skill that often serves as the basis for solving more complex geometric problems. The formula area = length × width simplifies this task. For this exercise, knowing the area and one dimension relationship simplified finding the other:

• Given: Area = 36 square meters
• Formula: Area = \(w \times (w + 3)\)

The equation \(w^2 + 3w - 36 = 0\) directly resulted from setting this equation with the given area. Solving it provided the width, which was then used to determine the length by substituting back into the expression for length. By rigorously using the area calculation, you can efficiently unveil unknown dimensions when some information is provided, demonstrating the practical application of area in geometry.