Problem 23 - 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. $$ l=\frac{x}{x-1} \text { and } w=\frac{3}{x-1} $$

Step-by-Step Solution

Answer

Perimeter: \( \frac{2(x+3)}{x-1} \); Area: \( \frac{3x}{(x-1)^2} \).

Step 1: Perimeter Formula

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

Step 2: Plug in Length and Width

Substitute the given expressions for length and width into the perimeter formula: \( P = 2 \left( \frac{x}{x-1} + \frac{3}{x-1} \right) \).

Step 3: Simplify the Expression

Add the fractions inside the parentheses: \( \frac{x}{x-1} + \frac{3}{x-1} = \frac{x+3}{x-1} \).

Step 4: Calculate the Perimeter

Distribute the 2 across the simplified fraction: \( P = 2 \times \frac{x+3}{x-1} = \frac{2(x+3)}{x-1} \).

Step 5: Area Formula

The area of a rectangle is given by \( A = l \times w \).

Step 6: Plug in Length and Width for Area

Substitute the given expressions for length and width to find the area: \( A = \left( \frac{x}{x-1} \right) \times \left( \frac{3}{x-1} \right) \).

Step 7: Simplify the Area Expression

Multiply the fractions: \( A = \frac{3x}{(x-1)^2} \).

Key Concepts

Perimeter of a Rectangle

To find the perimeter of a rectangle, you use a simple yet powerful formula: the sum of all its sides. The perimeter, denoted by \( P \), is calculated using the formula:

• \( P = 2(l + w) \)

This formula tells us to add the length \( l \) and the width \( w \), then multiply the sum by 2. This works because a rectangle has two pairs of equal sides. For example, if one side measures 3 units and the other 2 units, the perimeter is 10 units. It's like walking around the rectangle and counting each step you take.

Using this formula becomes essential when dimensions are given as algebraic expressions, like in our case. We are given:\( l=\frac{x}{x-1} \) and \( w=\frac{3}{x-1} \), which helps us find the perimeter in terms of the variable \( x \) by substituting and simplifying those expressions in the perimeter formula.

Area of a Rectangle

The area of a rectangle represents the amount of space it occupies. To calculate the area, you simply multiply its length by its width. The formula is:

• \( A = l \times w \)

Imagine laying out tiles to cover the rectangle without gaps, that's the area you're measuring. Each tile represents a unit square. So, if a rectangle has a length of 5 units and a width of 3 units, its area is 15 square units.

In this exercise, finding the area involves combining the given algebraic expressions for \( l \) and \( w \), which are \( \frac{x}{x-1} \) and \( \frac{3}{x-1} \) respectively. This requires multiplying two fractions, and as it turns out, the resulting expression \( A = \frac{3x}{(x-1)^2} \) reveals how the area changes with different values of \( x \).

Algebraic Expressions

Algebraic expressions are like math sentences that use variables, numbers, and operation signs to describe relationships and quantities. They allow us to express and manipulate mathematical ideas concisely. In our rectangle problem, both the length \( l \) and width \( w \) are given as fractions with variables:

• \( l = \frac{x}{x-1} \)
• \( w = \frac{3}{x-1} \)

These expressions tell us that as \( x \) changes, the dimensions of the rectangle change too. Through algebraic expressions, we can explore these changes systematically.

By substituting these variable expressions into the perimeter and area formulas, we gain insight into how these rectangles behave as \( x \) varies. It's like having a rulebook for how different scenarios unfold, without needing to test each one by hand.

Variable Manipulation

Variable manipulation is the art of rewriting and simplifying expressions to make them more manageable. It's how we solve algebra problems, like finding the perimeter and area of a rectangle here.

Consider the manipulation involved in computing the perimeter: after substituting the expressions \( l \) and \( w \), you need to add fractions:

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

This step is crucial because it consolidates the information into a simpler form, \( \frac{x+3}{x-1} \), which is easier to multiply by 2.

For the area, multiplication of fractions is key:

• \( \left( \frac{x}{x-1} \right) \times \left( \frac{3}{x-1} \right) = \frac{3x}{(x-1)^2} \)

This shows how algebraic manipulation helps streamline complex expressions, allowing them to be used efficiently in equations and problem solving.