Problem 17

Question

The length of a rectangle is three times its width. If the perimeter of the rectangle is 112 inches, find its length and width.

Step-by-Step Solution

Verified
Answer
The width is 14 inches and the length is 42 inches.
1Step 1: Understand the Problem
We have a rectangle where the length is three times the width, and we are given that the perimeter is 112 inches. We need to find both the length and the width of the rectangle.
2Step 2: Define Variables
Let the width of the rectangle be denoted as \( w \) inches. Then the length of the rectangle will be \( 3w \) inches, since the length is three times the width.
3Step 3: Write the Formula for Perimeter
The perimeter of a rectangle is calculated as \( 2 \times (\text{length} + \text{width}) \). Since we know the perimeter is 112 inches, we can substitute the length and width into the formula: \[ 2 \times (3w + w) = 112 \]
4Step 4: Simplify the Equation
Simplify the expression inside the parentheses: \[ 2 \times (4w) = 112 \] This simplifies to: \[ 8w = 112 \]
5Step 5: Solve for the Width
Divide both sides of the equation by 8 to find the width: \[ w = \frac{112}{8} = 14 \] Thus, the width of the rectangle is 14 inches.
6Step 6: Find the Length
Using the relation \( \text{length} = 3w \), substitute \( w = 14 \): \[ \text{length} = 3 \times 14 = 42 \] Thus, the length of the rectangle is 42 inches.

Key Concepts

Rectangle PerimeterVariable DefinitionProblem Solving Steps
Rectangle Perimeter
The perimeter of a rectangle is an essential concept in elementary algebra. To determine the perimeter, we need to calculate the total distance around the rectangle. This is done by adding up the lengths of all four sides. In mathematical terms, the formula for the perimeter, \( P \), of a rectangle is expressed as:
  • \( P = 2 \times (\text{length} + \text{width}) \)
When solving perimeter problems, it is important to know both the length and the width of the rectangle. If only the perimeter is given, as in the original exercise, additional relationships between the length and width (such as one being a multiple of the other) are key to finding both dimensions.
By understanding the formula, you can substitute the given values to find the unknown sides, making it a crucial step in solving rectangle perimeter problems.
Variable Definition
Defining variables is a fundamental step in approaching algebraic problems. In the exercise, we need to define a variable to represent the unknown dimension of the rectangle. Here's how it works:
  • Let the width of the rectangle be referred to as \( w \).
  • Since the length is three times the width, define the length in terms of \( w \) as \( 3w \).
Using variables in this way allows us to express unknown quantities with algebraic expressions. It simplifies the problem, enabling us to use equations to relate the known quantities and solve for the unknowns. This method of variable definition is essential in breaking down more complex problems into manageable parts, particularly in algebra.
Problem Solving Steps
Approaching math problems systematically ensures that we don't miss any critical steps. Let's break down the essential steps used in solving the rectangle perimeter problem:
  • Understand the problem: Identify what you're asked to find and what information is given. In this exercise, the known perimeter is 112 inches, and the relationship between length and width helps set the stage.
  • Define variables: Establish the variables to represent unknown values, which allows equations to be formed. Here, \( w \) represents the width.
  • Set up the equation: Based on the formula for the perimeter and the relation between length and width, build a mathematical equation. This involves substituting variables into known formulas.
  • Simplify and solve: By simplifying the equation, solve for the variable, finding the width first in this case.
  • Find other unknowns: Using the solved variable (width), calculate the length. This final step completes the solution.
Following these problem-solving steps ensures you solve algebraic problems efficiently and accurately. It fosters a structured approach that is vital for tackling a variety of mathematical problems.