Finding Rectangle Dimensions With Given Area And Length-Width Relationship

Understanding how to calculate the dimensions of a rectangle given its area and a relationship between its length and width is a fundamental concept in geometry. This article provides a comprehensive guide on how to solve such problems, focusing on a specific example where the area of a rectangle is 44 square yards, and the length is 3 yards less than twice the width. We will break down the problem step-by-step, explaining the underlying mathematical principles and providing clear, concise instructions.

Problem Statement

The area of a rectangle is given as 44 square yards ($yd^2$). The length of the rectangle is 3 yards less than twice the width. Our task is to find the dimensions of the rectangle, specifically its length and width.

Let's dive into the solution process, ensuring we cover all the necessary steps and concepts.

1. Define the Variables

In any mathematical problem, the first crucial step is to define the variables. This helps in translating the word problem into a mathematical equation. In our case, we have two unknowns: the length and the width of the rectangle.

• Let ${ w }$ represent the width of the rectangle in yards.
• Let ${ l }$ represent the length of the rectangle in yards.

Defining these variables is the foundation upon which we will build our equations and solve for the unknowns.

2. Translate the Given Information into Equations

Now that we have defined our variables, we need to translate the information given in the problem statement into mathematical equations. We have two pieces of information:

1. The area of the rectangle is 44 square yards.
2. The length of the rectangle is 3 yards less than twice the width.

Using the formula for the area of a rectangle, which is ${ \text{Area} = \text{length} \times \text{width} }$, we can write our first equation:

${ l \times w = 44 }$

This equation represents the area of the rectangle.

The second piece of information tells us that the length is 3 yards less than twice the width. We can translate this into the following equation:

${ l = 2w - 3 }$

This equation expresses the length in terms of the width. Now we have a system of two equations with two variables, which we can solve.

3. Solve the System of Equations

We have the following system of equations:

1. ${ l \times w = 44 }$
2. ${ l = 2w - 3 }$

A common method for solving such systems is substitution. We can substitute the expression for ${ l }$ from the second equation into the first equation. This will give us an equation in terms of ${ w }$ only.

Substituting ${ l = 2w - 3 }$ into ${ l \times w = 44 }$, we get:

${ (2w - 3)w = 44 }$

Expanding this equation, we obtain:

${ 2w^2 - 3w = 44 }$

To solve this quadratic equation, we need to set it equal to zero:

${ 2w^2 - 3w - 44 = 0 }$

Now we have a quadratic equation in the standard form ${ ax^2 + bx + c = 0 }$, where ${ a = 2 }$, ${ b = -3 }$, and ${ c = -44 }$. We can solve this equation by factoring, completing the square, or using the quadratic formula. In this case, factoring is a straightforward approach.

4. Factor the Quadratic Equation

We need to factor the quadratic equation:

${ 2w^2 - 3w - 44 = 0 }$

We are looking for two numbers that multiply to ${ 2 \times -44 = -88 }$ and add up to ${ -3 }$. These numbers are ${ -11 }$ and ${ 8 }$. So we can rewrite the middle term as:

${ 2w^2 - 11w + 8w - 44 = 0 }$

Now, we factor by grouping:

${ w(2w - 11) + 4(2w - 11) = 0 }$

${ (w + 4)(2w - 11) = 0 }$

Setting each factor equal to zero gives us two possible solutions for ${ w }$:

1. ${ w + 4 = 0 \implies w = -4 }$
2. ${ 2w - 11 = 0 \implies w = \frac{11}{2} = 5.5 }$

Since the width of a rectangle cannot be negative, we discard the solution ${ w = -4 }$. Therefore, the width of the rectangle is ${ w = 5.5 }$ yards.

5. Calculate the Length

Now that we have the width, we can find the length using the equation:

${ l = 2w - 3 }$

Substituting ${ w = 5.5 }$ into the equation, we get:

${ l = 2(5.5) - 3 }$

${ l = 11 - 3 }$

${ l = 8 }$

So, the length of the rectangle is 8 yards.

6. Verify the Solution

It's always a good practice to verify our solution to ensure that it satisfies the given conditions. We found that the width ${ w = 5.5 }$ yards and the length ${ l = 8 }$ yards. Let's check if these values satisfy the given conditions:

1. Area: ${ l \times w = 8 \times 5.5 = 44 }$ square yards, which matches the given area.
2. Length in terms of width: ${ l = 2w - 3 \implies 8 = 2(5.5) - 3 \implies 8 = 11 - 3 \implies 8 = 8 }$, which is also correct.

Since both conditions are satisfied, our solution is correct.

Final Answer

The dimensions of the rectangle are:

• Length: 8 yards
• Width: 5.5 yards

This step-by-step solution demonstrates how to solve problems involving the dimensions of rectangles, given their area and a relationship between their length and width. By defining variables, translating information into equations, solving the system of equations, and verifying the solution, we can confidently find the correct dimensions. Understanding these principles is essential for solving various geometric problems.

Additional Practice Problems

To solidify your understanding, here are a few additional practice problems:

1. Rectangle Area and Perimeter: The perimeter of a rectangle is 56 inches, and the length is 4 inches more than the width. Find the dimensions of the rectangle.
2. Rectangle Area and Diagonal: The area of a rectangle is 60 square feet, and the diagonal is 17 feet. Find the dimensions of the rectangle.
3. Rectangle with Changed Dimensions: The length of a rectangle is increased by 20%, and the width is decreased by 10%. If the new area is 108 square meters, and the original length was 12 meters, find the original width.

Solving these problems will give you further practice in applying the principles discussed in this article.

Conclusion

Mastering the methods to solve geometric problems, such as finding the dimensions of a rectangle, is crucial in mathematics. This article provided a detailed walkthrough, emphasizing each step from defining variables to verifying the solution. By following these steps and practicing with additional problems, you can enhance your problem-solving skills and confidently tackle similar challenges. Remember, the key to success in mathematics is understanding the fundamental concepts and applying them consistently.