The Length Of A Rectangular Room Is 2cm Longer Than The Width. What Should The Width Be For The Area To Be 15 Cm?

This article delves into the fascinating world of geometry, specifically focusing on solving a classic problem involving rectangles. We will explore how to determine the width of a rectangle when given its area and a relationship between its length and width. This is a fundamental concept in mathematics with practical applications in various fields, from architecture and engineering to everyday tasks like home improvement projects. Understanding this concept not only strengthens your mathematical skills but also enhances your problem-solving abilities in real-world scenarios. Let's embark on this journey to unravel the mysteries of rectangles and their dimensions.

Understanding the Problem

Before we dive into the solution, let's break down the problem statement. We are given a rectangle where the length is 2 cm greater than the width. Our goal is to find the width that will result in an area of 15 square centimeters. To approach this problem systematically, we will employ algebraic techniques. We'll represent the unknown width with a variable, express the length in terms of the width, and then use the formula for the area of a rectangle to form an equation. Solving this equation will reveal the desired width. This method demonstrates the power of algebra in translating word problems into mathematical expressions that can be readily solved. It's a powerful tool that allows us to tackle a wide range of geometrical challenges. This initial setup is crucial for a clear and efficient solution, laying the foundation for the subsequent algebraic manipulations. We will meticulously walk through each step, ensuring a comprehensive understanding of the process.

Setting up the Equation

Let's begin by assigning a variable to the unknown width. We'll use 'w' to represent the width of the rectangle in centimeters. Since the length is 2 cm greater than the width, we can express the length as 'w + 2' centimeters. The area of a rectangle is given by the product of its length and width. Therefore, we can write the equation as: Area = Length × Width, which translates to 15 = (w + 2) × w. This equation is a quadratic equation, which we will need to solve for 'w'. Understanding how to translate the word problem into this algebraic equation is a key skill in mathematical problem-solving. This step bridges the gap between the verbal description and the mathematical representation, making the problem amenable to algebraic techniques. By carefully defining our variables and applying the formula for the area of a rectangle, we have successfully set up the foundation for solving the problem. This methodical approach is essential for accuracy and clarity in mathematical calculations. The subsequent steps will involve manipulating this equation to isolate the variable and find its value.

Solving the Quadratic Equation

Now that we have our equation, 15 = (w + 2) × w, let's solve for 'w'. First, we expand the right side of the equation: 15 = w^2 + 2w. To solve this quadratic equation, we need to rearrange it into the standard form, which is ax^2 + bx + c = 0. Subtracting 15 from both sides, we get w^2 + 2w - 15 = 0. Next, we can factor the quadratic expression. We are looking for two numbers that multiply to -15 and add up to 2. These numbers are 5 and -3. So, we can factor the equation as (w + 5)(w - 3) = 0. Now, we set each factor equal to zero and solve for 'w': w + 5 = 0 or w - 3 = 0. This gives us two possible solutions: w = -5 or w = 3. However, since the width of a rectangle cannot be negative, we discard the solution w = -5. Therefore, the width of the rectangle is 3 cm. The ability to solve quadratic equations is a fundamental skill in algebra, and this example demonstrates its practical application in geometry. By systematically rearranging, factoring, and solving the equation, we have successfully determined the width of the rectangle. This process showcases the power of algebraic techniques in solving real-world problems. The next step will be to verify our solution and ensure it aligns with the given conditions.

Verifying the Solution

We found that the width of the rectangle is 3 cm. Let's verify if this solution satisfies the given conditions. If the width (w) is 3 cm, then the length (w + 2) is 3 + 2 = 5 cm. The area of the rectangle is length × width, which is 5 cm × 3 cm = 15 square centimeters. This matches the given area in the problem statement. Therefore, our solution is correct. Verifying the solution is a crucial step in problem-solving. It ensures that the answer obtained is not only mathematically correct but also makes sense in the context of the problem. This step reinforces the understanding of the problem and the solution process. By confirming that the calculated dimensions result in the given area, we can be confident in our answer. This rigorous approach to problem-solving is essential for accuracy and building a strong foundation in mathematics. It also highlights the interconnectedness of different mathematical concepts, such as algebra and geometry.

Final Answer and Conclusion

Therefore, the width of the rectangle must be 3 cm for its area to be 15 square centimeters. This problem demonstrates the application of algebraic techniques in solving geometrical problems. By setting up an equation based on the given information and solving it, we were able to find the unknown width of the rectangle. This process involved understanding the relationship between the length, width, and area of a rectangle, as well as the ability to manipulate algebraic expressions and solve quadratic equations. The solution not only provides the answer to the specific problem but also illustrates the general approach to solving similar problems. The key takeaway is the importance of translating word problems into mathematical expressions and using algebraic tools to find the solution. This skill is invaluable in various fields, including mathematics, science, and engineering. By mastering these techniques, one can confidently tackle a wide range of quantitative challenges. The final answer of 3 cm for the width of the rectangle is a testament to the power of mathematical reasoning and problem-solving skills.

The dimensions of a rectangle, namely its length and width, are fundamental properties that dictate its size and shape. These dimensions are crucial in various applications, from calculating the area and perimeter of a rectangular space to designing structures and objects. Understanding the relationship between length, width, and other properties like area is essential for solving geometrical problems and making informed decisions in practical scenarios. In this section, we will delve deeper into the significance of rectangular dimensions and how they influence various calculations and applications. This exploration will provide a solid foundation for tackling more complex problems involving rectangles and other geometrical shapes.

Importance of Length and Width

The length and width of a rectangle are the two primary measurements that define its size. The length is typically the longer side, while the width is the shorter side. However, in some contexts, these terms may be used interchangeably. These dimensions are not just abstract measurements; they have tangible implications in the real world. For instance, when planning a room layout, the length and width of the room determine the amount of furniture that can fit and the overall spatial arrangement. In construction, the dimensions of a rectangular plot of land dictate the size and shape of the building that can be erected. Similarly, in manufacturing, the dimensions of a rectangular component influence its functionality and compatibility with other parts. The accurate measurement and consideration of length and width are therefore paramount in numerous fields. These dimensions serve as the building blocks for calculating other important properties, such as area and perimeter, which are crucial for various design and planning purposes. A thorough understanding of length and width is therefore indispensable for anyone working with rectangular shapes or spaces.

Relationship to Area

The area of a rectangle is directly determined by its length and width. As we discussed earlier, the formula for the area of a rectangle is Area = Length × Width. This simple equation encapsulates a fundamental relationship between these three quantities. If we know the length and width, we can easily calculate the area. Conversely, if we know the area and one of the dimensions (either length or width), we can determine the other dimension by rearranging the formula. This relationship is not only mathematically significant but also practically useful. For example, if we want to cover a rectangular floor with tiles, we need to know the area of the floor to calculate the number of tiles required. Similarly, if we have a fixed amount of material to build a rectangular enclosure, we need to understand the relationship between length, width, and area to maximize the enclosed space. The interplay between length, width, and area is therefore a cornerstone of geometrical calculations and has wide-ranging applications in various fields. Mastering this relationship is crucial for solving problems involving rectangles and other geometrical shapes.

Practice Problems

To solidify your understanding of solving problems involving the dimensions and area of rectangles, let's explore some practice problems. These problems will test your ability to apply the concepts and techniques we have discussed in this article. Remember to break down each problem into smaller steps, identify the given information, set up equations, solve for the unknowns, and verify your solutions. Practice is key to mastering any mathematical concept, and these problems will provide you with valuable experience in tackling different scenarios involving rectangles.

Problem-Solving Strategies

When faced with a problem involving the dimensions and area of a rectangle, it's helpful to have a systematic approach. Start by carefully reading the problem statement and identifying the knowns and unknowns. Draw a diagram of the rectangle and label the dimensions with variables if necessary. Write down the relevant formulas, such as the formula for the area of a rectangle. If there is a relationship between the length and width, express one dimension in terms of the other. Set up an equation based on the given information and solve for the unknown variables. Finally, verify your solution by plugging the values back into the original equation or problem statement. This structured approach will help you tackle complex problems with confidence and accuracy.

Example 1

A rectangular garden has a length that is 3 meters longer than its width. If the area of the garden is 40 square meters, find the width and length of the garden.

Solution:

Let 'w' be the width of the garden in meters. Then the length is 'w + 3' meters. The area is given by Length × Width, so we have the equation: (w + 3) × w = 40. Expanding the equation, we get w^2 + 3w = 40. Rearranging into standard form, we have w^2 + 3w - 40 = 0. Factoring the quadratic, we get (w + 8)(w - 5) = 0. The possible solutions are w = -8 and w = 5. Since the width cannot be negative, we have w = 5 meters. Therefore, the length is w + 3 = 5 + 3 = 8 meters. The width of the garden is 5 meters, and the length is 8 meters.

Example 2

The perimeter of a rectangle is 24 cm, and its length is twice its width. Find the dimensions of the rectangle.

Solution:

Let 'w' be the width of the rectangle in centimeters. Then the length is '2w' centimeters. The perimeter of a rectangle is given by 2 × (Length + Width), so we have the equation: 2 × (2w + w) = 24. Simplifying the equation, we get 2 × (3w) = 24, which leads to 6w = 24. Dividing both sides by 6, we get w = 4 cm. Therefore, the length is 2w = 2 × 4 = 8 cm. The width of the rectangle is 4 cm, and the length is 8 cm.

Example 3

A rectangle has an area of 36 square inches. If the length is 9 inches, what is the width?

Solution:

The area of a rectangle is given by Length × Width. We know the area is 36 square inches and the length is 9 inches. Let 'w' be the width in inches. We have the equation: 9 × w = 36. Dividing both sides by 9, we get w = 4 inches. Therefore, the width of the rectangle is 4 inches.

These examples illustrate the application of the problem-solving strategies discussed earlier. By carefully setting up equations and using algebraic techniques, we can successfully solve a variety of problems involving rectangles. These practical examples further solidify the understanding of rectangular dimensions.