Problem 7
Question
Find the area and perimeter of the rectangle in terms of the width \(W\). The width \(W\) is half the length.
Step-by-Step Solution
Verified Answer
Area = \(2W^2\); Perimeter = \(6W\).
1Step 1: Define the length in terms of width
According to the problem, the width \( W \) is half the length. Thus, the length \( L \) can be expressed as \( L = 2W \).
2Step 2: Calculate the area of the rectangle
The area of a rectangle is given by the formula \( \text{Area} = \text{Length} \times \text{Width} \). Substituting the expression for the length, the area becomes \( \text{Area} = 2W \times W = 2W^2 \).
3Step 3: Calculate the perimeter of the rectangle
The perimeter of a rectangle is given by the formula \( \text{Perimeter} = 2L + 2W \). Substituting the expression for the length, the perimeter becomes \( \text{Perimeter} = 2(2W) + 2W = 4W + 2W = 6W \).
4Step 4: Conclusion
The area and perimeter of the rectangle in terms of the width \( W \) are \( 2W^2 \) and \( 6W \) respectively.
Key Concepts
Understanding the Area of a RectangleCalculating the Perimeter of a RectangleUsing Algebraic Expressions with Rectangles
Understanding the Area of a Rectangle
The area of a rectangle is the amount of space the rectangle covers on a flat surface. Imagine coloring the entire rectangle with your favorite color. That's the area! To find this, you multiply the length by the width. When we're told that width \( W \) is half of the length, we look for a way to express the length using \( W \) as well. Here, the length \( L \) is double the width. Mathematically, that's written as \( L = 2W \). Using the area formula \( \text{Area} = \text{Length} \times \text{Width} \):
- Substitute the length with \( 2W \).
- Compute the area: \( \text{Area} = 2W \times W = 2W^2 \).
Calculating the Perimeter of a Rectangle
The perimeter is like the belt around the waist of the rectangle. It's the total distance you travel if you walk around the edges once. To find the perimeter, we add all the sides' lengths. The formula is \( \text{Perimeter} = 2L + 2W \). Again, with \( L = 2W \), we can substitute into the perimeter formula:
- Replace \( L \) with \( 2W \): \( \text{Perimeter} = 2(2W) + 2W \).
- Compute step by step: \( \text{Perimeter} = 4W + 2W = 6W \).
Using Algebraic Expressions with Rectangles
Algebraic expressions are like recipes for calculations: they show how to compute things step by step using symbols and numbers. In this exercise, you express both the area and the perimeter of a rectangle using the variable \( W \), which represents the width.Understanding algebraic expressions:
- They allow you to generalize rectangle properties using variables.
- They make calculations easier and flexible for different values of \( W \).
Other exercises in this chapter
Problem 7
Factor out the greatest common factor:. \(5 x^{4}-15 x^{3}+15 x^{2}\)
View solution Problem 7
Simplify the expression. $$ \frac{x+3}{2 x^{2}+5 x-3} $$
View solution Problem 8
Combine like terms whenever possible. $$-y^{2}-\frac{1}{2} y^{2}$$
View solution Problem 8
Simplify the expression. Assume that all variables are positive. $$ \sqrt[3]{\frac{x}{8}} $$
View solution