Problem 21
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. $$ l=2 x \text { and } w=\frac{1}{x} $$
Step-by-Step Solution
Verified Answer
Perimeter: \( 4x + \frac{2}{x} \), Area: \( 2 \).
1Step 1: Understand the Problem
We are given the length \( l = 2x \) and the width \( w = \frac{1}{x} \) of a rectangle. Our task is to express the perimeter and area of this rectangle in terms of the variable \( x \).
2Step 2: Find the Perimeter Formula
The formula for the perimeter \( P \) of a rectangle is given by \( P = 2(l + w) \). Substitute the given expressions for length and width: \( P = 2(2x + \frac{1}{x}) \).
3Step 3: Simplify the Perimeter Expression
Simplify the expression for the perimeter: \( P = 2(2x + \frac{1}{x}) = 4x + \frac{2}{x} \). Thus, the perimeter in terms of \( x \) is \( 4x + \frac{2}{x} \).
4Step 4: Find the Area Formula
The formula for the area \( A \) of a rectangle is \( A = l imes w \). Substitute the expressions for length and width: \( A = (2x) \left( \frac{1}{x} \right) \).
5Step 5: Simplify the Area Expression
Simplify the expression for the area: \( A = 2x \cdot \frac{1}{x} = 2 \). Thus, the area in terms of \( x \) is \( 2 \).
Key Concepts
Rectangle GeometryPerimeter CalculationArea Calculation
Rectangle Geometry
Understanding the basics of rectangle geometry is an important part of algebra and geometry. A rectangle is a four-sided figure with opposite sides that are both equal in length and parallel. This symmetry and shape make it easy to calculate its dimensions such as perimeter and area.
Key properties of a rectangle include:
Key properties of a rectangle include:
- Four right angles, each measuring 90 degrees.
- Opposite sides that are equal in length (length and width).
- Parallel opposite sides.
Perimeter Calculation
Calculating the perimeter of a rectangle involves summing up the lengths of all its four sides. In rectangles, this is simplified by the fact that opposite sides are equal. The general formula for finding the perimeter of a rectangle is: \[ P = 2(l + w) \] Where \( l \) represents the length and \( w \) the width. In the context of variable expressions, substituting the expressions into this formula gives us the perimeter in terms of that variable.
In the given exercise, where the length is \( l = 2x \) and the width is \( w = \frac{1}{x} \), the perimeter becomes:\[ P = 2(2x + \frac{1}{x}) = 4x + \frac{2}{x} \]
This result shows how algebraic expressions can be used to calculate and represent geometric dimensions even when direct numerical values are not available.
In the given exercise, where the length is \( l = 2x \) and the width is \( w = \frac{1}{x} \), the perimeter becomes:\[ P = 2(2x + \frac{1}{x}) = 4x + \frac{2}{x} \]
This result shows how algebraic expressions can be used to calculate and represent geometric dimensions even when direct numerical values are not available.
Area Calculation
The area of a rectangle gives us the total amount of space contained within its four sides. It is calculated by multiplying the length by the width, which gives a formula of: \[ A = l \times w \]
When dealing with algebraic expressions, this formula can be adapted by substituting the given expressions for length and width. In this context, for a length of \( l = 2x \) and a width of \( w = \frac{1}{x} \), the area becomes:\[ A = (2x) \left( \frac{1}{x} \right) = 2 \]
This means the area remains constant at 2, regardless of the value of \( x \). In problems like this, simplifying algebraic expressions to reach a numerical conclusion is crucial, as it helps students understand how variable expressions affect geometric dimensions.
When dealing with algebraic expressions, this formula can be adapted by substituting the given expressions for length and width. In this context, for a length of \( l = 2x \) and a width of \( w = \frac{1}{x} \), the area becomes:\[ A = (2x) \left( \frac{1}{x} \right) = 2 \]
This means the area remains constant at 2, regardless of the value of \( x \). In problems like this, simplifying algebraic expressions to reach a numerical conclusion is crucial, as it helps students understand how variable expressions affect geometric dimensions.
Other exercises in this chapter
Problem 20
In \(13-22,\) write each decimal as a common fraction. $$ 0.5 \overline{7} $$
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Simplify each expression. In each case, list any values of the variables for which the fractions are not defined. \(\frac{3}{2 x}-\frac{1+\frac{1}{x}}{x+1}\)
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Last week, the ratio of the number of hours that Joseph worked to the number of hours that Nicole worked was \(2 : 3 .\) This week Joseph worked 4 hours more th
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In \(13-24,\) divide and express each quotient in simplest form. In each case, list any values of the variables for which the fractions are not defined. $$ \fra
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