Solving For Painting Length Using Perimeter And Algebraic Equations

Jul 11, 2025 by ADMIN 68 views

Framing a Rectangular Painting: Solving for Length with Equations

Introduction

In this article, we will delve into a practical problem involving geometry and algebra. We'll explore how to determine the dimensions of a rectangular painting given the amount of framing material used and the relationship between its length and width. This exercise combines mathematical concepts with real-world applications, making it an engaging way to understand problem-solving techniques. We'll break down the problem step-by-step, formulating an equation and solving it to find the length of the painting. This exploration will not only reinforce your understanding of algebraic principles but also enhance your ability to apply them in everyday scenarios. Join us as we unravel this mathematical puzzle and discover the elegant solution.

Problem Statement: Framing a Rectangular Painting

Lisa is framing a rectangular painting. The length of the painting is three inches more than twice its width. Lisa uses 30 inches of framing material to frame the painting. The primary question we aim to answer is: What is the length of the painting? To solve this, we will first need to establish an equation that represents the given information and then solve that equation for the unknown variable. This problem is a classic example of how algebra can be used to model and solve real-world geometric problems. The key to unlocking the solution lies in understanding the relationship between the perimeter of a rectangle and its dimensions, and then translating the word problem into a mathematical equation. Let's dive into the details and begin our journey towards the solution.

Setting Up the Equation

To solve this problem effectively, we must first translate the given information into a mathematical equation. This involves identifying the unknowns and establishing relationships between them. In this scenario, the unknowns are the width and the length of the rectangular painting. Let's denote the width of the painting as 'w' inches. According to the problem statement, the length of the painting is three inches more than twice its width. We can express this relationship algebraically as length 'l' = 2w + 3. Now, we know that Lisa uses 30 inches of framing material, which corresponds to the perimeter of the rectangular painting. The formula for the perimeter (P) of a rectangle is given by P = 2l + 2w. We can substitute the given perimeter value (30 inches) and the expression for the length (2w + 3) into the perimeter formula. This substitution allows us to create an equation with a single variable, which we can then solve to find the width of the painting. The equation will be instrumental in determining the dimensions of the painting and ultimately answering the question about its length. By carefully setting up the equation, we pave the way for a systematic and accurate solution.

Solving the Equation

Now that we have established the equation, the next step is to solve it for the unknown variable, which will give us the width of the painting. Let's recap the equation we derived: 30 = 2(2w + 3) + 2w. To solve this equation, we'll follow the standard algebraic procedure. First, we'll distribute the 2 across the terms inside the parentheses: 30 = 4w + 6 + 2w. Next, we combine like terms on the right side of the equation: 30 = 6w + 6. To isolate the term with the variable, we subtract 6 from both sides of the equation: 30 - 6 = 6w + 6 - 6, which simplifies to 24 = 6w. Finally, to solve for w, we divide both sides of the equation by 6: 24 / 6 = 6w / 6, which gives us w = 4. This means the width of the painting is 4 inches. However, our ultimate goal is to find the length of the painting. We already established the relationship between the length and the width as l = 2w + 3. Now that we know the value of w, we can substitute it into this equation to find the length. This step demonstrates the power of algebraic manipulation in solving geometric problems, allowing us to systematically arrive at the solution.

Calculating the Length

With the width of the painting determined to be 4 inches, we are now in a position to calculate the length. Recall the relationship we established earlier: the length (l) is three inches more than twice the width (w). Mathematically, this is represented as l = 2w + 3. To find the length, we simply substitute the value of w (which is 4 inches) into this equation. Thus, l = 2(4) + 3. Performing the multiplication, we get l = 8 + 3. Adding these values together, we find that l = 11 inches. Therefore, the length of the painting is 11 inches. This calculation completes the mathematical process of solving the problem, providing a concrete answer to the initial question. It showcases how algebraic equations can be used to model and solve geometric problems, demonstrating the practical application of mathematical principles in real-world scenarios. The final answer not only provides the length of the painting but also reinforces the importance of careful algebraic manipulation in problem-solving.

Verification of the Solution

Before concluding, it is crucial to verify our solution to ensure its accuracy. We have determined that the width of the painting is 4 inches and the length is 11 inches. To verify, we can use the perimeter formula, P = 2l + 2w, and check if these dimensions result in the given framing material length of 30 inches. Substituting the values we found, we get P = 2(11) + 2(4). Performing the multiplications, we have P = 22 + 8. Adding these values, we find that P = 30 inches, which matches the given amount of framing material. This confirms that our solution is correct. Additionally, we can revisit the problem statement and check if the length (11 inches) is indeed three inches more than twice the width (4 inches). Twice the width is 2 * 4 = 8 inches, and adding 3 inches to that gives us 11 inches, which is the length we calculated. This double verification process reinforces our confidence in the accuracy of the solution. Verifying the solution is an essential step in problem-solving, ensuring that the answer is not only mathematically correct but also makes sense in the context of the problem.

Conclusion

In summary, we successfully determined the length of the rectangular painting by translating the word problem into an algebraic equation and solving it systematically. We started by defining the variables for the width and length, establishing the relationship between them based on the given information. We then used the perimeter formula to create an equation representing the total framing material used. By solving this equation, we found the width of the painting to be 4 inches. Subsequently, we used the relationship between the length and width to calculate the length, which turned out to be 11 inches. Finally, we verified our solution by plugging the calculated dimensions back into the perimeter formula, confirming that they matched the given framing material length. This exercise demonstrates the power of algebraic problem-solving in real-world contexts. By breaking down the problem into smaller, manageable steps, we were able to arrive at a precise and accurate solution. This approach not only enhances our understanding of mathematical concepts but also improves our ability to apply them in practical situations. The process of setting up an equation, solving it, and verifying the solution is a valuable skill that can be applied to a wide range of problems in mathematics and beyond.

Keywords and SEO Optimization

This article addresses the problem of finding the dimensions of a rectangular painting given its perimeter and the relationship between its length and width. To optimize for search engines and ensure readability, the following keywords have been strategically incorporated:

Rectangular painting: This is the central object of the problem, and the phrase is used throughout the article.
Framing material: This refers to the perimeter of the painting and is a key element in setting up the equation.
Length and width: These are the dimensions we are trying to find, and they are mentioned frequently in the context of the problem and solution.
Equation: This term highlights the algebraic approach used to solve the problem.
Perimeter: This geometric concept is crucial for setting up the equation and verifying the solution.
Algebraic problem-solving: This encompasses the overall methodology used in the article.

These keywords have been naturally integrated into the content to improve search engine visibility while maintaining a clear and engaging narrative for the reader. The article also includes a step-by-step explanation of the problem-solving process, making it accessible and informative for anyone interested in mathematics and practical applications of algebra.