Problem 8 - Algebra and Trigonometry

Question

Express the given quantity in terms of the indicated variable. The perimeter (in \(\mathrm{cm} )\) of a rectangle that is 5 \(\mathrm{cm}\) longer than it is wide; \(w=\) width of the rectangle (in \(\mathrm{cm} )\)

Step-by-Step Solution

Answer

The perimeter is \(4w + 10\) cm.

Step 1: Understanding the Problem

To find an expression for the perimeter of a rectangle, we need to express it in terms of its width, given that the rectangle is 5 cm longer than it is wide.

Step 2: Identifying Rectangle Dimensions

Given that the rectangle's length is 5 cm longer than its width, express the length as: \(\text{Length} = w + 5\), where \(w\) is the width.

Step 3: Defining the Perimeter Formula

The perimeter \(P\) of a rectangle is given by the formula: \(P = 2(\text{Length} + \text{Width})\).

Step 4: Substituting Dimensions into the Formula

Replace the length in the perimeter formula: \(P = 2((w + 5) + w)\).

Step 5: Simplifying the Expression

Combine like terms in the expression: \(P = 2(2w + 5)\). Simplify further by distributing the 2: \(P = 4w + 10\).

Step 6: Final Expression for Perimeter

The expression for the perimeter of the rectangle in terms of its width is \(4w + 10\).

Key Concepts

Expression Simplification

Simplifying expressions in mathematics refers to the process of making them easier to understand and work with. For example, when we discuss the perimeter of a rectangle, we might start with a complicated expression that includes multiple terms and variables. The goal is to combine like terms and reduce the expression to the simplest form possible.

In our problem, we start with the expression for the perimeter:

• \[ P = 2((w + 5) + w) \]

Expanding the terms inside the brackets, it becomes helpful to spot these like terms. In this case:

• \[ w + w = 2w \]
• \[ 5 \] is a constant.

Thus, the expression becomes:

• \[ 2(2w + 5) \]

Simplifying further by distributing the 2 throughout each term results in:

• \[ 4w + 10 \]

The expression is now as simple as possible. It is direct and straightforward, clearly showing the relationship between the perimeter and the width of the rectangle.

Rectangle Properties

A rectangle is a four-sided polygon with some unique properties. Understanding these properties is essential when solving problems related to rectangles.

• **Opposite Sides Equality**: In a rectangle, opposite sides are equal in length. For our problem, this translates to having two widths and two lengths.
• **Right Angles**: Every angle in a rectangle is a right angle (90 degrees). This feature guarantees that all sides are perpendicular to each other.
• **Perimeter Formula**: The perimeter is the total distance around the rectangle. It is calculated with the formula: \[ P = 2( ext{Length} + ext{Width}) \]

These properties help define relations in problems involving rectangles such as calculating perimeter, area, or understanding the geometrical layout. In our exercise, knowing that the rectangle's length is '5 cm longer than its width' aligns with these properties, making it easier to set up the algebraic expression.

Algebraic Expression

Algebraic expressions are used to represent mathematical quantities. They allow us to write general rules or properties in terms of variables and constants. For the problem we have, the goal is to express the perimeter in terms of the rectangle's width using an algebraic expression.

In this case, the length of the rectangle is expressed as \( w + 5 \), where \( w \) is a variable representing the width. This forms the foundation of our algebraic expression for the perimeter. As variables can take on different values, our expression remains general until specific values are substituted.

We use this flexibility to rewrite the perimeter formula in terms of \( w \). By substituting the expression for the length into our perimeter formula and simplifying it accordingly, we arrive at:

• Final Algebraic Expression for Perimeter: \[ P = 4w + 10 \]

This expression neatly represents the perimeter of any rectangle where the length is 5 cm more than its width, demonstrating the power and usefulness of algebra in handling geometrical problems.