Problem 22 - Algebra 2 and Trigonometry

Question

In \(21-24,\) the length and width of a rectangle are expressed in terms of a variable. a. Express each perimeter in terms of the variable. b. Express each area in terms of the variable. $$ l=3 x+3 \text { and } w=\frac{1}{3} $$

Step-by-Step Solution

Answer

Perimeter: \( P = 6x + \frac{20}{3} \), Area: \( A = x + 1 \).

Step 1: Understand the Problem

We are given the length \( l = 3x + 3 \) and the width \( w = \frac{1}{3} \) of a rectangle. Our task is to express the perimeter and area in terms of \( x \).

Step 2: Recall the Perimeter Formula

The formula for the perimeter \( P \) of a rectangle is given by \( P = 2(l + w) \).

Step 3: Substitute Values into Perimeter Formula

Substitute the given expressions for \( l \) and \( w \) into the formula:\[ P = 2\left((3x + 3) + \frac{1}{3}\right) \].

Step 4: Simplify the Expression for Perimeter

Simplify inside the parentheses: \\( 3x + 3 + \frac{1}{3} = 3x + \frac{9}{3} + \frac{1}{3} = 3x + \frac{10}{3} \).Now, simplify the perimeter formula: \\( P = 2\left(3x + \frac{10}{3}\right) = 6x + \frac{20}{3} \).

Step 5: Recall the Area Formula

The formula for the area \( A \) of a rectangle is \( A = l \times w \).

Step 6: Substitute Values into Area Formula

Substitute the given expressions for \( l \) and \( w \):\[ A = (3x + 3) \times \frac{1}{3} \].

Step 7: Simplify the Expression for Area

Distribute \( \frac{1}{3} \) across the terms \\( A = \frac{1}{3} \times 3x + \frac{1}{3} \times 3 = x + 1 \).

Step 8: Write Down the Final Expressions

The perimeter of the rectangle in terms of \( x \) is \( P = 6x + \frac{20}{3} \), and the area is \( A = x + 1 \).

Key Concepts

Perimeter

The perimeter of a rectangle is the total distance around the outside of the rectangle. It can be calculated using the lengths of the sides. For rectangles, the formula to find the perimeter is:

• \( P = 2(l + w) \)

where \( l \) is the length and \( w \) is the width. In this exercise, the length \( l \) is given by \( 3x + 3 \) and the width \( w \) is \( \frac{1}{3} \). Adding these together forms the expression inside the parentheses:

• \( 3x + 3 + \frac{1}{3} = 3x + \frac{9}{3} + \frac{1}{3} = 3x + \frac{10}{3} \)

Then, using the perimeter formula, substitute this expression:

• \( P = 2(3x + \frac{10}{3}) \)

Simplifying gives:

• \( P = 6x + \frac{20}{3} \)

This expression shows how the perimeter changes as the variable \( x \) changes.

Area

The area of a rectangle is the amount of space enclosed within its sides and is calculated by multiplying the length by the width. The formula used is:

• \( A = l \times w \)

In this scenario, substituting the given expressions, \( l = 3x + 3 \) and \( w = \frac{1}{3} \), into the formula gives:

• \( A = (3x + 3) \times \frac{1}{3} \)

By distributing the \( \frac{1}{3} \) across the terms in the expression:

• \( A = \frac{1}{3} \times 3x + \frac{1}{3} \times 3 \)
• \( A = x + 1 \)

This simplified expression \( A = x + 1 \) shows how the area will vary depending on the values of \( x \).

Variable Expressions

Variable expressions allow us to write relationships using letters, often called variables, to represent numbers. This makes them useful in algebra. In this exercise, the variable \( x \) is used to express both the length and width of a rectangle. The expression for length is \( l = 3x + 3 \), showing that it is directly affected by \( x \). Similarly, the width has a constant term as \( w = \frac{1}{3} \).Variable expressions convert a fixed formula into something flexible that can adapt to different scenarios by changing the value of \( x \). This lets you calculate outcomes like the perimeter and area for any number that \( x \) might take. By substituting \( x \) into these expressions, it becomes clear how both the perimeter and area change in response to the value of \( x \) shifting.This approach helps in visualizing how algebra uses mathematical operations to express real-world scenarios, offering insights into relationships between different quantities such as dimensions of shapes.