Solving For Dimensions Of A Rectangular Room With A Fixed Area

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In the realm of geometry, rectangles hold a fundamental place, their properties and dimensions often serving as the basis for various calculations and problem-solving scenarios. One such scenario involves determining the dimensions of a rectangular room given its area and a relationship between its length and width. In this article, we embark on a mathematical exploration to unravel the dimensions of a rectangular room with an area of 750 square feet, where the width is 5 feet less than the length. We will delve into the process of formulating equations that can be used to solve for the length of the room, represented by the variable 'y'.

Formulating the Equations

To begin our exploration, let's define the variables involved. We are given that the length of the room is represented by 'y', and the width is 5 feet less than the length. Therefore, we can express the width as 'y - 5'. The area of a rectangle is calculated by multiplying its length and width, so we have the equation:

  • Area = Length × Width
  • 750 = y(y - 5)

This equation forms the foundation for our quest to determine the length of the room. To solve for 'y', we need to manipulate this equation into a form that allows us to isolate the variable.

Expanding and Rearranging the Equation

Our next step involves expanding the equation and rearranging it into a standard quadratic form. Expanding the equation, we get:

  • 750 = y² - 5y

To bring the equation into standard quadratic form (ax² + bx + c = 0), we subtract 750 from both sides:

  • 0 = y² - 5y - 750

This quadratic equation now presents us with a familiar challenge – finding the values of 'y' that satisfy the equation. We can approach this challenge using various methods, including factoring, the quadratic formula, or completing the square.

Factoring the Quadratic Equation

Factoring involves expressing the quadratic equation as a product of two binomials. To factor the equation y² - 5y - 750 = 0, we need to find two numbers that multiply to -750 and add up to -5. These numbers are -30 and 25. Therefore, we can rewrite the equation as:

  • (y - 30)(y + 25) = 0

This factored form provides us with two possible solutions for 'y':

  • y - 30 = 0 or y + 25 = 0
  • y = 30 or y = -25

Since the length of a room cannot be negative, we discard the solution y = -25. Therefore, the length of the room is y = 30 feet.

Verifying the Solution

To ensure the accuracy of our solution, we can substitute y = 30 back into the original equation:

  • 750 = y(y - 5)
  • 750 = 30(30 - 5)
  • 750 = 30(25)
  • 750 = 750

The equation holds true, confirming that our solution of y = 30 feet is correct.

Exploring Alternative Equations

While we have successfully solved for the length of the room using factoring, it's worth exploring other equations that could also be used to arrive at the same solution. Let's consider the following equations:

  1. (y + 25)(y - 30) = 0

    This equation is the factored form of the quadratic equation we derived earlier. As we discussed, this equation directly leads to the solutions y = 30 and y = -25, with y = 30 being the valid solution for the length of the room.

  2. y² - 5y = 750

    This equation represents the expanded form of the original equation before we rearranged it into standard quadratic form. It is a valid equation that can be used to solve for 'y', although it may require further manipulation to isolate the variable.

  3. y² - 5y - 750 = 0

    This equation is the standard quadratic form of the equation, which we obtained by rearranging the expanded form. It is a crucial step in solving the equation using methods like the quadratic formula or completing the square.

  4. y² - 5 = 750

    This equation is incorrect. It does not accurately represent the relationship between the length, width, and area of the rectangular room. The term '-5y' is missing, which accounts for the difference between the length and width.

  5. y(y - 5) = 750

    This equation is the original equation we formulated based on the given information. It directly represents the relationship between the length, width, and area of the room and is a valid equation for solving for 'y'.

Conclusion

In this mathematical exploration, we successfully unraveled the dimensions of a rectangular room with an area of 750 square feet, where the width is 5 feet less than the length. We formulated equations that could be used to solve for the length of the room, and we identified the correct equations among a set of options. The equations (y + 25)(y - 30) = 0, y² - 5y - 750 = 0, and y(y - 5) = 750 are all valid equations that can be used to determine the length of the room. This exploration highlights the power of mathematical equations in representing real-world scenarios and solving for unknown quantities.

By understanding the relationships between geometric properties and algebraic equations, we can effectively tackle problems involving shapes, dimensions, and areas. This knowledge is not only valuable in academic settings but also in practical applications such as architecture, engineering, and design.

In summary, the journey to find the length of the rectangular room involved:

  • Defining variables: Representing the length as 'y' and the width as 'y - 5'.
  • Formulating the equation: Expressing the area as the product of length and width: 750 = y(y - 5).
  • Expanding and rearranging: Transforming the equation into standard quadratic form: y² - 5y - 750 = 0.
  • Solving the equation: Using factoring to find the solutions y = 30 and y = -25.
  • Choosing the valid solution: Discarding the negative solution and accepting y = 30 feet as the length.
  • Verifying the solution: Substituting y = 30 back into the original equation to confirm its accuracy.
  • Exploring alternative equations: Identifying other valid equations that could be used to solve for 'y'.

This comprehensive approach showcases the interconnectedness of mathematical concepts and their application in solving real-world problems. As we continue to explore the realm of mathematics, we will encounter more such scenarios where our understanding of equations and geometric principles will empower us to unravel complex challenges.

Choosing the Correct Equations

In the original problem, we were tasked with selecting three correct equations that could be used to solve for 'y', the length of the room. Based on our exploration, the three correct answers are:

  1. (y + 25)(y - 30) = 0
  2. y² - 5y - 750 = 0
  3. y(y - 5) = 750

These equations represent different forms of the same mathematical relationship, each offering a pathway to finding the value of 'y'. By understanding the underlying principles and applying appropriate techniques, we can confidently navigate the world of equations and unlock the solutions to various problems.

Practical Applications and Further Exploration

The problem of finding the dimensions of a rectangular room is not just a theoretical exercise. It has practical applications in various fields, such as:

  • Architecture and Interior Design: Architects and interior designers use these calculations to determine the optimal layout of rooms, ensuring efficient use of space and comfortable living environments.
  • Construction and Engineering: Construction workers and engineers rely on accurate dimensional calculations to build structures that are safe, stable, and meet the required specifications.
  • Real Estate and Property Management: Real estate agents and property managers use area calculations to determine property values and rental rates.

Furthermore, this problem serves as a stepping stone to more complex geometric challenges. We can extend our exploration by considering rooms with irregular shapes, calculating volumes of three-dimensional spaces, and investigating the relationships between different geometric figures.

By embracing the power of mathematics and its applications, we can unlock a deeper understanding of the world around us and develop the skills necessary to tackle a wide range of challenges.

Keywords

  • Rectangular Room Dimensions
  • Area Calculation
  • Quadratic Equations
  • Factoring Quadratics
  • Problem-Solving in Geometry
  • Mathematical Applications
  • Length and Width Relationship
  • Real-World Applications of Geometry
  • Architectural Calculations
  • Engineering Mathematics