Solving For Rectangle Width Given Perimeter And Length Relationship

In the realm of mathematics, geometry problems often present intriguing challenges. One common type involves finding the dimensions of a rectangle given its perimeter and a relationship between its length and width. This article delves into a specific instance of this problem, where we're tasked with determining the width of a rectangle. Our main keyword here is understanding the problem itself. The problem states that the perimeter of a rectangle is 38 inches, and the length is 3 inches more than the width. Our goal is to find the width. This type of problem requires us to translate word problems into algebraic equations and then solve them. To solve this effectively, we need a firm grasp of the formula for the perimeter of a rectangle, which is P = 2l + 2w, where P represents the perimeter, l represents the length, and w represents the width. Understanding this formula is crucial because it forms the foundation for setting up our equation. The given information provides us with two key pieces of data: the perimeter (38 inches) and the relationship between the length and width (length is 3 inches more than the width). This relationship can be expressed algebraically as l = w + 3. Now, we have all the necessary components to begin solving the problem. We can substitute the given information into the perimeter formula and solve for the unknown width. The beauty of this type of problem lies in its ability to demonstrate the practical application of algebraic concepts. It shows how we can use mathematical tools to solve real-world scenarios involving shapes and measurements. By carefully analyzing the problem statement and translating it into mathematical expressions, we can systematically arrive at the solution. In the following sections, we will walk through the steps involved in solving this particular problem, providing a clear and concise explanation of each step. We will see how the combination of geometric understanding and algebraic manipulation allows us to decipher the dimensions of the rectangle. So, let's embark on this mathematical journey and unravel the mystery of the rectangle's width. We'll break down the problem into manageable parts, making it easy to follow along and grasp the underlying principles.

Setting up the equation is a crucial step in solving any word problem, especially in mathematics. This is where we translate the given information into a mathematical form that we can work with. In our rectangle problem, the key is to represent the unknowns (length and width) using variables and then form an equation based on the given perimeter and the relationship between length and width. Let's denote the width of the rectangle as 'w'. Since the length is 3 inches more than the width, we can represent the length as 'w + 3'. Now, we know that the perimeter of a rectangle is given by the formula P = 2l + 2w. We are given that the perimeter is 38 inches. So, we can substitute P = 38 into the formula. We also know that l = w + 3. So, we can substitute this expression for 'l' into the perimeter formula as well. This gives us the equation 38 = 2(w + 3) + 2w. This equation now represents the problem in a mathematical form. It captures the relationship between the perimeter, length, and width of the rectangle, as described in the problem statement. The next step is to solve this equation for 'w', which will give us the width of the rectangle. Before we proceed with solving the equation, let's take a moment to appreciate the importance of this step. Setting up the equation correctly is often the most challenging part of solving word problems. It requires careful reading and understanding of the problem statement, as well as the ability to translate the information into mathematical symbols and expressions. A well-set-up equation makes the subsequent steps of solving for the unknown much easier. On the other hand, an incorrectly set-up equation will lead to the wrong answer, regardless of how accurately we perform the algebraic manipulations. So, it's crucial to take our time and ensure that we have correctly captured the essence of the problem in our equation. In our case, we have successfully translated the given information into the equation 38 = 2(w + 3) + 2w. This equation is the key to unlocking the solution to our problem. In the next section, we will walk through the steps involved in solving this equation and finding the value of 'w', which represents the width of the rectangle.

Now that we have our equation, 38 = 2(w + 3) + 2w, the next step is to solve for the width, 'w'. This involves applying algebraic principles to isolate 'w' on one side of the equation. First, we need to simplify the equation by distributing the 2 in the term 2(w + 3). This gives us 38 = 2w + 6 + 2w. Next, we combine like terms on the right side of the equation. We have 2w + 2w, which simplifies to 4w. So, our equation becomes 38 = 4w + 6. Now, we want to isolate the term with 'w', which is 4w. To do this, we subtract 6 from both sides of the equation. This gives us 38 - 6 = 4w + 6 - 6, which simplifies to 32 = 4w. Finally, to solve for 'w', we divide both sides of the equation by 4. This gives us 32 / 4 = 4w / 4, which simplifies to w = 8. So, we have found that the width of the rectangle is 8 inches. But we're not done yet! It's always a good practice to check our answer to make sure it makes sense in the context of the original problem. We found that the width is 8 inches. The problem stated that the length is 3 inches more than the width, so the length would be 8 + 3 = 11 inches. Now, let's calculate the perimeter using these values. The perimeter is given by P = 2l + 2w. Substituting our values, we get P = 2(11) + 2(8) = 22 + 16 = 38 inches. This matches the perimeter given in the problem statement, so our answer is consistent with the given information. We have successfully solved for the width of the rectangle and verified our solution. This step-by-step approach demonstrates the power of algebra in solving geometric problems. By carefully applying algebraic principles, we can systematically isolate the unknown variable and find its value. This process also highlights the importance of checking our answer to ensure its validity. In the next section, we will summarize our findings and discuss the significance of this problem-solving approach.

In conclusion, we have successfully navigated the rectangle perimeter problem, demonstrating a powerful blend of geometric understanding and algebraic techniques. Our main keyword here is summarizing the entire problem-solving process and the strategies employed. We started by carefully analyzing the problem statement, identifying the given information (perimeter = 38 inches, length = width + 3 inches), and recognizing the unknown we needed to find (the width). The key to solving this problem was setting up the equation correctly. We translated the word problem into a mathematical equation using the formula for the perimeter of a rectangle, P = 2l + 2w, and the relationship between the length and width, l = w + 3. This resulted in the equation 38 = 2(w + 3) + 2w. Once we had the equation, we systematically applied algebraic principles to solve for the width, 'w'. We distributed, combined like terms, isolated the variable term, and finally divided to find the value of w. This process yielded the solution w = 8 inches. But our journey didn't end there. We recognized the importance of verifying our solution to ensure its accuracy. We plugged the value of w back into the original problem and calculated the perimeter using the found width and the derived length. The calculated perimeter matched the given perimeter, confirming the validity of our solution. This problem-solving approach highlights several important strategies that are applicable to a wide range of mathematical problems. First, careful reading and understanding of the problem statement are crucial. Identifying the given information and the unknown is the first step towards a solution. Second, translating word problems into mathematical equations is a key skill. This involves representing unknowns with variables and expressing relationships using mathematical symbols and operations. Third, applying algebraic principles systematically is essential for solving equations. This includes distributing, combining like terms, isolating variables, and performing operations on both sides of the equation to maintain balance. Finally, verifying the solution is a critical step. This helps to ensure that the answer is correct and makes sense in the context of the problem. By mastering these strategies, we can confidently tackle a variety of mathematical challenges. This rectangle perimeter problem serves as a valuable example of how mathematics can be used to solve real-world problems involving shapes, measurements, and relationships. It also underscores the importance of a methodical and systematic approach to problem-solving.

Now that we've successfully solved the rectangle perimeter problem, let's consider how we can further explore the concept and extend our understanding. This is a main keyword for encouraging deeper learning and application of the concepts. One way to do this is to modify the problem slightly and see how the solution changes. For example, we could change the relationship between the length and width. Instead of the length being 3 inches more than the width, we could say the length is twice the width, or the length is 5 inches less than the width. How would these changes affect the equation we set up and the final solution? Another modification we could make is to change the given perimeter. What if the perimeter was 40 inches, or 30 inches? How would this change the width of the rectangle? By experimenting with these variations, we can gain a deeper understanding of how the perimeter, length, and width of a rectangle are related. We can also consider more complex problems involving rectangles. For example, we could be given the area and the perimeter of a rectangle and asked to find the length and width. This type of problem requires us to use two equations, one for the area and one for the perimeter, and solve them simultaneously. Another extension could involve introducing other geometric shapes. For example, we could be given a figure that is composed of a rectangle and a triangle, and asked to find the perimeter or area of the entire figure. This would require us to apply our knowledge of the properties of both rectangles and triangles. We could also explore real-world applications of rectangle perimeter problems. For example, we might need to calculate the amount of fencing needed to enclose a rectangular garden, or the amount of trim needed to go around a rectangular window. These types of problems demonstrate the practical relevance of the mathematical concepts we've been learning. Finally, we could delve deeper into the underlying algebraic principles that we used to solve the problem. For example, we could explore different methods for solving linear equations, or we could investigate the properties of geometric shapes in more detail. By pushing our understanding beyond the specific problem we solved, we can develop a more comprehensive grasp of mathematics and its applications. This continuous exploration and extension of concepts is key to becoming a proficient problem solver. So, let's continue to challenge ourselves and discover the fascinating world of mathematics.