Rectangular Perimeter and Area: Problem Solving and Applications
Have you ever faced a geometry problem that seemed a bit daunting at first glance? Let's dive into a problem that involves calculating the perimeter and area of a rectangle, which is a common question in geometry. This article will walk you through the problem-solving process, providing clear explanations and step-by-step solutions. We'll look at five different examples, all dealing with the relationship between a rectangle's length, width, and area.
Problem 1: Given the Relation between Length and Width
Problem Statement: The length of a rectangle is six times its width. If the area of the rectangle is 384 ft2, what is its perimeter?
Solution:
Let the width of the rectangle be w feet. Then the length l can be expressed as:
l 6w
The area A of the rectangle is given by:
A l × w
Substituting the expression for length into the area formula, we have:
A 6w × w 6w2
Given that the area is 384 ft2, we can set up the equation:
6w2 384
Dividing both sides by 6:
w2 384 / 6 64
Take the square root of both sides:
w √64 8 ft
Now, we can find the length:
l 6w 6 × 8 48 ft
Finally, we calculate the perimeter P of the rectangle using the formula:
P 2l 2w
Substituting the values of length and width:
P 2 × 48 2 × 8 96 16 112 ft
Thus, the perimeter of the rectangle is:
112 ft
Problem 2: Another Example with Variables
Problem Statement: Let the length be 5 times the breadth. If the area is 125 ft2, find the perimeter.
Solution:
Let the breadth be w feet. Then the length can be expressed as:
L 5w
The area of the rectangle is given by:
A L × w 5w × w 5w2
Given that the area is 125 ft2, we can set up the equation:
5w2 125
Dividing both sides by 5:
w2 125 / 5 25
Take the square root of both sides:
w √25 5 ft
Now, we can find the length:
L 5 × 5 25 ft
Next, we calculate the perimeter P of the rectangle:
P 2L 2w 2 × 25 2 × 5 50 10 60 ft
Thus, the perimeter of the rectangle is:
60 ft
Problem 3: Using Algebraic Expressions
Problem Statement: If the length of a rectangle is 5 times its width, and the area is 125 ft2, what is the perimeter?
Solution:
Let the breadth be w feet. Then the length can be expressed as:
L 5w
Using the area formula:
A L × w 5w × w 5w2
Given that the area is 125 ft2, we set up the equation:
5w2 125
Dividing both sides by 5:
w2 125 / 5 25
Take the square root of both sides:
w √25 5 ft
Now, we can find the length:
L 5 × 5 25 ft
Calculate the perimeter:
P 2L 2w 2 × 25 2 × 5 50 10 60 ft
Thus, the perimeter of the rectangle is:
60 ft
Problem 4: Working with Squared Variables
Problem Statement: If the length of a rectangle is 5 times its width, and the area is 405 in2, what is its perimeter?
Solution:
Let the width be w inches. Then the length can be expressed as:
L 5w
The area of the rectangle is given by:
A L × w 5w × w 5w2
Given that the area is 405 in2, we set up the equation:
5w2 405
Dividing both sides by 5:
w2 405 / 5 81
Take the square root of both sides:
w √81 9 in
Now, we can find the length:
L 5 × 9 45 in
Calculate the perimeter:
P 2L 2w 2 × 45 2 × 9 90 18 108 in
Thus, the perimeter of the rectangle is:
108 in
Problem 5: Applying Geometric Principles
Problem Statement: The length of a rectangle is 5 times its width, and the area is 405 in2. What is the perimeter?
Solution:
Let the width be w inches. Then the length can be expressed as:
L 5w
The area of the rectangle is given by:
A L × w 5w × w 5w2
Given that the area is 405 in2, we set up the equation:
5w2 405
Dividing both sides by 5:
w2 405 / 5 81
Take the square root of both sides:
w √81 9 in
Now, we can find the length:
L 5 × 9 45 in
Calculate the perimeter:
P 2L 2w 2 × 45 2 × 9 90 18 108 in
Thus, the perimeter of the rectangle is:
108 in
In conclusion, understanding the relationship between a rectangle's length, width, and area can be explored through various examples. These problems demonstrate that the key steps involve expressing the variables, setting up equations based on the given conditions, solving for one variable, and then calculating the other. The process is consistent across all five problems, making it easier to tackle similar questions in the future. The perimeter of a rectangle and the area of a rectangle are fundamental concepts in geometry, essential for problem-solving in both academic and practical contexts.