The Area Of A Rectangle Is 33 Ft², And The Length Of The Rectangle Is 5 Ft Less Than Twice The Width. Find The Dimensions Of The Rectangle. Length: _____ FtWidth: _____ Ft
=====================================================
Introduction
In this problem, we are given the area of a rectangle and a relationship between its length and width. We need to find the dimensions of the rectangle, which are the length and width. The area of a rectangle is given by the formula: Area = Length × Width. We are also given that the length of the rectangle is 5 ft less than twice the width.
Problem Statement
The area of a rectangle is 33 ft², and the length of the rectangle is 5 ft less than twice the width. We need to find the dimensions of the rectangle, which are the length and width.
Step 1: Define the Variables
Let's define the variables:
- L = Length of the rectangle
- W = Width of the rectangle
Step 2: Write the Equation
The area of the rectangle is given by the formula: Area = Length × Width. We are given that the area is 33 ft², so we can write the equation:
L × W = 33
We are also given that the length of the rectangle is 5 ft less than twice the width, so we can write another equation:
L = 2W - 5
Step 3: Substitute the Second Equation into the First Equation
We can substitute the second equation into the first equation to get:
(2W - 5) × W = 33
Step 4: Expand and Simplify the Equation
We can expand and simplify the equation to get:
2W² - 5W = 33
Step 5: Rearrange the Equation
We can rearrange the equation to get:
2W² - 5W - 33 = 0
Step 6: Solve the Quadratic Equation
We can solve the quadratic equation using the quadratic formula:
W = (-b ± √(b² - 4ac)) / 2a
In this case, a = 2, b = -5, and c = -33. Plugging these values into the formula, we get:
W = (5 ± √((-5)² - 4(2)(-33))) / 2(2)
W = (5 ± √(25 + 264)) / 4
W = (5 ± √289) / 4
W = (5 ± 17) / 4
We have two possible solutions for W:
W = (5 + 17) / 4 = 22 / 4 = 5.5
W = (5 - 17) / 4 = -12 / 4 = -3
Since the width cannot be negative, we discard the solution W = -3.
Step 7: Find the Length
Now that we have found the width, we can find the length using the equation:
L = 2W - 5
Plugging in W = 5.5, we get:
L = 2(5.5) - 5
L = 11 - 5
L = 6
Step 8: Check the Solution
We can check our solution by plugging the values of L and W back into the original equations. We get:
L × W = 6 × 5.5 = 33
L = 2W - 5 = 2(5.5) - 5 = 6
Our solution checks out!
Conclusion
In this problem, we were given the area of a rectangle and a relationship between its length and width. We used algebraic techniques to find the dimensions of the rectangle, which are the length and width. We found that the width is 5.5 ft and the length is 6 ft.
Final Answer
Length: 6 ft Width: 5.5 ft
Discussion
This problem is a classic example of a quadratic equation. We used the quadratic formula to solve the equation and found two possible solutions for the width. We discarded the negative solution and kept the positive solution. We then used the equation for the length to find the length of the rectangle. Our solution checks out, and we have found the dimensions of the rectangle.
Related Problems
- Finding the dimensions of a rectangle given the area and a relationship between the length and width.
- Solving quadratic equations using the quadratic formula.
- Using algebraic techniques to solve problems involving geometry.
References
================================
Q: What is the formula for the area of a rectangle?
A: The formula for the area of a rectangle is: Area = Length × Width.
Q: How do I find the dimensions of a rectangle given the area and a relationship between the length and width?
A: To find the dimensions of a rectangle given the area and a relationship between the length and width, you can use algebraic techniques. First, write the equation for the area of the rectangle: Area = Length × Width. Then, write the equation for the relationship between the length and width. Finally, substitute the second equation into the first equation and solve for the width. Once you have found the width, you can find the length using the equation for the relationship between the length and width.
Q: What is a quadratic equation?
A: A quadratic equation is an equation of the form: ax² + bx + c = 0, where a, b, and c are constants, and x is the variable. Quadratic equations can be solved using the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a.
Q: How do I solve a quadratic equation using the quadratic formula?
A: To solve a quadratic equation using the quadratic formula, you need to plug in the values of a, b, and c into the formula. Then, simplify the expression under the square root and solve for x.
Q: What is the quadratic formula?
A: The quadratic formula is: x = (-b ± √(b² - 4ac)) / 2a.
Q: How do I find the dimensions of a rectangle given the area and a relationship between the length and width if the relationship is not linear?
A: If the relationship between the length and width is not linear, you may need to use more advanced algebraic techniques, such as substitution or elimination. You can also use graphing techniques to find the solution.
Q: Can I use a calculator to solve a quadratic equation?
A: Yes, you can use a calculator to solve a quadratic equation. Most calculators have a quadratic formula function that you can use to solve the equation.
Q: What are some common mistakes to avoid when solving quadratic equations?
A: Some common mistakes to avoid when solving quadratic equations include:
- Not simplifying the expression under the square root
- Not plugging in the correct values of a, b, and c into the quadratic formula
- Not solving for x correctly
- Not checking the solution to make sure it is valid
Q: How do I check my solution to a quadratic equation?
A: To check your solution to a quadratic equation, you can plug the values of x back into the original equation and simplify. If the equation is true, then your solution is correct.
Q: What are some real-world applications of quadratic equations?
A: Quadratic equations have many real-world applications, including:
- Finding the dimensions of a rectangle or a square
- Modeling the trajectory of a projectile
- Finding the maximum or minimum value of a function
- Solving problems involving geometry and trigonometry
Related Questions
- How do I find the dimensions of a rectangle given the area and a relationship between the length and width?
- What is a quadratic equation?
- How do I solve a quadratic equation using the quadratic formula?
- What are some common mistakes to avoid when solving quadratic equations?
- How do I check my solution to a quadratic equation?
Related Articles
- The Area of a Rectangle: Finding the Dimensions
- Quadratic Equations: A Beginner's Guide
- Algebraic Techniques: A Guide to Solving Equations