Problem 47 - Algebra for College Students

Question

Solve each of Problems 47– 62 by setting up and solving an appropriate algebraic equation. Suppose that the length of a certain rectangle is \(2 \mathrm{me}-\) ters less than four times its width. The perimeter of the rectangle is 56 meters. Find the length and width of the rectangle.

Step-by-Step Solution

Answer

The width is 6 meters and the length is 22 meters.

Step 1: Set Up the Equation for Perimeter

Let the width of the rectangle be \( W \). According to the problem, the length \( L \) is given by \( L = 4W - 2 \). The formula for the perimeter \( P \) of a rectangle is \( P = 2L + 2W \). Since the perimeter is 56 meters, we have the equation: \[ 2(4W - 2) + 2W = 56 \].

Step 2: Simplify the Equation

Distribute and simplify the terms in the equation: \[ 8W - 4 + 2W = 56 \]. Combine like terms to get \[ 10W - 4 = 56 \].

Step 3: Solve for the Width

Add 4 to both sides of the equation to isolate the term with \( W \): \[ 10W = 60 \]. Divide both sides by 10 to find the width: \[ W = 6 \].

Step 4: Find the Length Using Width

Now that the width \( W \) is known, substitute \( W = 6 \) back into the expression for the length: \[ L = 4W - 2 \]. Substitute and calculate: \[ L = 4(6) - 2 = 24 - 2 = 22 \].

Key Concepts

Rectangle Perimeter

To tackle problems involving a rectangle's perimeter, it's essential to first understand what perimeter means. The perimeter of a rectangle is simply the total distance around the shape, which can be calculated by adding the length of all its sides. A rectangle has two lengths and two widths. Hence, the formula for the perimeter is given by:

• Perimeter (\( P \)) = 2 times the Length (\( L \)) plus 2 times the Width (\( W \)).
• Mathematically, it's expressed as \( P = 2L + 2W \).

Understanding this formula helps not only in calculating a given perimeter but also in solving problems where other dimensions of the rectangle are unknown. For instance, if you know the perimeter but not the dimensions, you can use this formula as a basis for setting up an equation, which is crucial in perimeter problem-solving. Breaking down the equation can thus lead to finding unknown quantities like the length or width.

Algebraic Equations

Algebraic equations serve as powerful tools in determining unknown variables in geometric problems, especially in finding dimensions of shapes based on given conditions. In our example, we are tasked with finding the length and width of a rectangle. Given that the length is related to the width by a certain formula, we can set up an equation to solve for the width.
First, represent the width by a variable, often denoted as \( W \). Then, express the length in terms of this width, such as \( L = 4W - 2 \).

• Use the perimeter formula \( P = 2L + 2W \) and substitute the expression for \( L \).
• This substitution yields a single equation in terms of \( W \).
• Simplify and rearrange algebraic terms to solve for \( W \).

Through this process, the abstract problem of finding the rectangle's dimensions becomes a neat exercise in solving algebraic equations. By systematically isolating and calculating, you can arrive at the needed values.

Width and Length Determination

After understanding how to set up and simplify algebraic equations, the next logical step is determining the dimensions such as width and length. Once the width \( W \) is known, in this case as \( W = 6 \), you can quickly determine the length with the expression derived earlier. \( L = 4W - 2 \).
Using the known width:

• Substitute \( W = 6 \) into the equation for the length.
• Calculate to find \( L = 4(6) - 2 \).
• The calculation results in \( L = 22 \).

Now, both the width and the length of the rectangle are determined, making the problem-solving process complete. By following a logical sequence of substituting known values into the equation, this approach makes it straightforward to derive all necessary dimensions of the rectangle, providing a complete solution to the problem.