Problem 24 - 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. $$ l=\frac{x}{x+1} \text { and } w=\frac{x}{x+2} $$

Step-by-Step Solution

Answer

Perimeter: \( \frac{4x^2 + 6x}{(x+1)(x+2)} \); Area: \( \frac{x^2}{(x+1)(x+2)} \).

Step 1: Perimeter Formula

The perimeter of a rectangle is calculated using the formula: \( P = 2l + 2w \). This formula adds double the length and double the width to get the total perimeter.

Step 2: Substitute Length and Width in Perimeter Formula

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

Step 3: Simplify the Perimeter Expression

Simplify the expression by distributing the 2 and combining like terms:\[ P = \frac{2x}{x+1} + \frac{2x}{x+2} \]

Step 4: Find a Common Denominator for Perimeter

To add the fractions, find a common denominator, which in this case is \((x+1)(x+2)\).Rewrite each fraction:\[ P = \frac{2x(x+2)}{(x+1)(x+2)} + \frac{2x(x+1)}{(x+1)(x+2)} \]

Step 5: Combine the Perimeter Over the Common Denominator

Add the numerators over the common denominator:\[ P = \frac{2x^2 + 4x + 2x^2 + 2x}{(x+1)(x+2)} \]Combine like terms:\[ P = \frac{4x^2 + 6x}{(x+1)(x+2)} \]

Step 6: Area Formula

The area of a rectangle is calculated using the formula: \( A = l \times w \). This involves multiplying the length by the width.

Step 7: Substitute Length and Width in Area Formula

Substitute the given expressions for length and width into the area formula.\[ A = \left(\frac{x}{x+1}\right) \times \left(\frac{x}{x+2}\right) \]

Step 8: Multiply the Length and Width

Multiply the numerators and the denominators separately:\[ A = \frac{x \times x}{(x+1)(x+2)} = \frac{x^2}{(x+1)(x+2)} \]

Step 9: Final Expressions

The perimeter of the rectangle is expressed as \( P = \frac{4x^2 + 6x}{(x+1)(x+2)} \) and the area is expressed as \( A = \frac{x^2}{(x+1)(x+2)} \).

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle is like the fence surrounding a garden. It represents the total length around the rectangle, including all sides. To calculate the perimeter, we use the formula:

• \( P = 2l + 2w \)

Here, \( l \) is the length and \( w \) is the width of the rectangle. By multiplying both the length and the width by 2 and adding them together, we get the complete measure of the boundary.
In the exercise, the length \( l \) is given as \( \frac{x}{x+1} \) and the width \( w \) is \( \frac{x}{x+2} \).
By substituting these into the perimeter formula, the perimeter \( P \) in terms of \( x \) becomes:

• \( P = 2\left(\frac{x}{x+1}\right) + 2\left(\frac{x}{x+2}\right) \)

This formula will help us understand how the perimeter changes with different values of \( x \).

Area of a Rectangle

The area of a rectangle tells us how much surface the rectangle covers. Imagine it as the amount of paint needed to fill the surface.
We calculate the area using the formula:

• \( A = l \times w \)

By multiplying the length (\( l \)) by the width (\( w \)), we determine the total space inside the rectangular boundary.
For our problem, substituting the length \( \frac{x}{x+1} \) and width \( \frac{x}{x+2} \) into the area formula gives us:

• \( A = \left(\frac{x}{x+1}\right) \times \left(\frac{x}{x+2}\right) \)

This expression simplifies to \( \frac{x^2}{(x+1)(x+2)} \), showing the dependency on \( x \). Understanding this helps in visualizing how the area varies with different \( x \) values.

Rational Expressions

Rational expressions are like fractions but involve polynomials. Think of them as a ratio of two algebraic expressions where the numerator and the denominator contain variables.
In our exercise, the length and width are rational expressions:

• \( l = \frac{x}{x+1} \)
• \( w = \frac{x}{x+2} \)

These expressions can behave differently compared to simple numbers. They require special attention when performing operations like addition, subtraction, multiplication, and division.
We'll often see rational expressions in algebra, and with practice, these become much easier to work with.

Common Denominator

A common denominator is a shared multiple of denominators in a set of fractions, essential for adding or subtracting them. Just like finding a common ground in a debate, it helps align different parts to interact seamlessly.
For instance, when dealing with the rational expressions in the exercise's perimeter equation:

• \( \frac{2x}{x+1} + \frac{2x}{x+2} \)

We need a common denominator to add these fractions. Here, it's \( (x+1)(x+2) \). By rewriting each fraction using this common ground, we can easily combine them:

• \( \frac{2x(x+2)}{(x+1)(x+2)} + \frac{2x(x+1)}{(x+1)(x+2)} \)

Once they share the denominator, adding the numerators combines the fractions smoothly. Understanding this concept is crucial for simplifying algebraic expressions and solving equations efficiently.