Rectangle Width Calculation Length 7 Cm Longer Than Width Perimeter 120 Cm

Understanding the Rectangle and its Properties

When dealing with geometric shapes, a rectangle is a fundamental figure to understand. In this specific problem, we delve into a rectangle where its length is intricately tied to its width, creating a unique scenario for us to explore. To effectively tackle this, let’s first define what a rectangle is. A rectangle, in its essence, is a quadrilateral – a four-sided polygon – with the distinct characteristic of having all its angles as right angles (90 degrees). This attribute immediately gives rise to some key properties. The opposite sides of a rectangle are not only parallel but also congruent, meaning they have the same length. This leads us to the two primary dimensions of a rectangle: its length and its width. The length typically refers to the longer side, while the width refers to the shorter side. However, in the case of a square, where all sides are equal, the distinction blurs, but the principles remain the same. The perimeter of a rectangle, another crucial concept, is the total distance around its boundary. It's essentially the sum of all its sides. For a rectangle, this translates to twice the sum of its length and its width. Mathematically, we express this as Perimeter = 2 * (Length + Width). This formula is the key to unlocking the solution to many rectangle-related problems, including the one at hand. Now, we introduce the specific condition given in our problem: the length of the rectangle is 7 cm longer than its width. This introduces a relationship between the two dimensions, which we can express algebraically. If we denote the width as 'w', then the length can be represented as 'w + 7'. This algebraic representation is a powerful tool as it allows us to translate the word problem into a mathematical equation, paving the way for a solution. Furthermore, we are given that the perimeter of this rectangle is 120 cm. This piece of information is critical as it provides a concrete value that we can use in our perimeter formula. Combining this with the relationship we established between the length and width, we can set up an equation that involves only one variable, the width 'w'. Solving this equation will lead us to the width of the rectangle, which is the ultimate goal of our problem. In the subsequent sections, we will delve into the algebraic formulation and the step-by-step solution to unravel the dimensions of this rectangle.

Setting Up the Equation: Connecting Length, Width, and Perimeter

To effectively solve this geometric problem, the initial crucial step involves translating the word problem into a concrete algebraic equation. This equation will serve as the backbone of our solution, allowing us to systematically determine the width of the rectangle. We begin by defining our variables. Let's denote the width of the rectangle as 'w', a standard practice in algebraic problem-solving. Given that the length of the rectangle is 7 cm longer than its width, we can express the length in terms of 'w' as 'w + 7'. This is a pivotal step as it establishes a direct relationship between the two dimensions of the rectangle, encapsulating the core information provided in the problem statement. Next, we recall the formula for the perimeter of a rectangle: Perimeter = 2 * (Length + Width). This formula is a fundamental concept in geometry and forms the basis of our equation. We are given that the perimeter of the rectangle is 120 cm. Now, we substitute the expressions for Length and Width, which we defined earlier in terms of 'w', into the perimeter formula. This yields the equation: 120 = 2 * ((w + 7) + w). This equation is the heart of our solution strategy. It mathematically represents the relationship between the perimeter, length, and width of the rectangle, incorporating the specific condition that the length is 7 cm longer than the width. The equation now involves only one variable, 'w', which represents the width of the rectangle. This makes it solvable using standard algebraic techniques. The next step involves simplifying this equation. We begin by distributing the 2 on the right side of the equation, which gives us 120 = 2 * (2w + 7). Further simplification involves multiplying the 2 into the parentheses, resulting in 120 = 4w + 14. This simplified equation is now in a more manageable form. The goal is to isolate 'w' on one side of the equation, which will give us the value of the width. To do this, we will first subtract 14 from both sides of the equation. This maintains the balance of the equation while moving us closer to isolating 'w'. The subtraction results in 120 - 14 = 4w + 14 - 14, which simplifies to 106 = 4w. We are now just one step away from finding the value of 'w'. In the next section, we will complete the process by isolating 'w' and determining the width of the rectangle.

Solving for the Width: The Algebraic Steps

Having established the equation 106 = 4w, the next crucial step is to solve for 'w', which represents the width of the rectangle. This involves using basic algebraic principles to isolate 'w' on one side of the equation. Our current equation, 106 = 4w, indicates that 106 is equal to 4 times the width. To find the value of 'w', we need to reverse this operation. The reverse of multiplication is division. Therefore, we will divide both sides of the equation by 4. This maintains the balance of the equation, ensuring that we are performing a mathematically sound operation. Dividing both sides by 4 gives us 106 / 4 = (4w) / 4. On the right side, the 4 in the numerator and the 4 in the denominator cancel each other out, leaving us with just 'w'. On the left side, we perform the division of 106 by 4. This division results in 26.5. Therefore, our equation now reads 26.5 = w. This tells us that the width of the rectangle, represented by 'w', is 26.5 cm. It's important to include the units (cm in this case) in our answer to provide context to the numerical value. We have now successfully determined the width of the rectangle. However, to ensure the accuracy of our solution and to provide a complete answer to the problem, it is beneficial to also find the length of the rectangle and verify that our solution satisfies the given conditions. We recall that the length of the rectangle is given by 'w + 7'. Now that we know 'w' is 26.5 cm, we can substitute this value into the expression for the length. This gives us Length = 26.5 + 7. Performing the addition, we find that the length is 33.5 cm. Now that we have both the width (26.5 cm) and the length (33.5 cm), we can verify our solution by checking if the perimeter calculated using these dimensions matches the given perimeter of 120 cm. We use the perimeter formula: Perimeter = 2 * (Length + Width). Substituting our values, we get Perimeter = 2 * (33.5 + 26.5). Adding the length and width gives us 33.5 + 26.5 = 60. Multiplying this by 2, we get Perimeter = 2 * 60 = 120 cm. This confirms that our calculated dimensions satisfy the given perimeter, validating our solution. In conclusion, the width of the rectangle is 26.5 cm. In the next section, we will discuss the implications of our solution and highlight the key steps in the problem-solving process.

Final Answer: Width of the Rectangle and Solution Verification

Having meticulously worked through the problem, we have arrived at the final answer: the width of the rectangle is 26.5 cm. This conclusion is the culmination of a series of logical steps, from translating the word problem into an algebraic equation to solving for the unknown variable. To recap, we were given a rectangle with the condition that its length is 7 cm longer than its width, and its perimeter is 120 cm. Our task was to find the width of this rectangle. We began by denoting the width as 'w' and expressing the length in terms of 'w' as 'w + 7'. We then utilized the formula for the perimeter of a rectangle, Perimeter = 2 * (Length + Width), and substituted our expressions for length and width to create the equation 120 = 2 * ((w + 7) + w). This equation was the key to unlocking the solution. We simplified the equation, combining like terms and isolating 'w' on one side. This involved distributing the 2, combining the 'w' terms, and then performing algebraic manipulations to isolate 'w'. Specifically, we subtracted 14 from both sides of the equation and then divided both sides by 4. This process led us to the solution w = 26.5 cm. It is crucial to include the units (cm in this case) to provide a complete and meaningful answer. However, our problem-solving process didn't end with just finding the width. We went a step further to verify our solution. This is a crucial step in any mathematical problem-solving endeavor. To verify, we calculated the length of the rectangle using the width we found. Since the length is 'w + 7', we substituted w = 26.5 cm to find Length = 33.5 cm. With both the width and length known, we calculated the perimeter using the formula Perimeter = 2 * (Length + Width). Substituting our values, we found the perimeter to be 120 cm, which matches the given perimeter in the problem. This verification step confirms the accuracy of our solution. In summary, the width of the rectangle is 26.5 cm, and the length is 33.5 cm. These dimensions satisfy the conditions given in the problem, namely that the length is 7 cm longer than the width and the perimeter is 120 cm. This problem exemplifies the power of algebraic methods in solving geometric problems. By translating the word problem into a mathematical equation, we were able to systematically determine the unknown dimension of the rectangle.

Importance of Understanding Perimeter and Algebraic Equations

This exercise underscores the importance of understanding fundamental geometric concepts like perimeter and the ability to translate word problems into algebraic equations. The concept of perimeter, as we've seen, is not just about memorizing a formula; it's about understanding the total distance around a shape. This understanding is crucial in various real-world applications, from fencing a garden to designing a room layout. In this specific problem, the perimeter served as a constraint, a piece of information that tied the length and width together. Without understanding the concept of perimeter and its relationship to the dimensions of a rectangle, we wouldn't have been able to set up the initial equation. The ability to translate word problems into algebraic equations is another critical skill highlighted by this exercise. Word problems often present information in a narrative format, and it's up to us to extract the relevant mathematical relationships and express them in a symbolic form. This involves identifying the unknowns, assigning variables, and formulating equations that represent the given conditions. In our problem, we translated the statement "the length is 7 cm longer than the width" into the algebraic expression "length = w + 7". This translation was a crucial step in setting up the equation that led to the solution. Furthermore, this problem reinforces the importance of algebraic manipulation skills. Once we had the equation 120 = 2 * ((w + 7) + w), we needed to simplify it and isolate the variable 'w'. This involved applying the distributive property, combining like terms, and using inverse operations to solve for 'w'. These algebraic skills are fundamental not only in mathematics but also in many other fields, including science, engineering, and economics. Beyond the specific solution, this problem illustrates a general approach to problem-solving. First, we carefully read and understood the problem, identifying the given information and the unknown we needed to find. Second, we translated the word problem into a mathematical model, using variables and equations to represent the relationships between the quantities. Third, we solved the equation using algebraic techniques. Fourth, we verified our solution to ensure its accuracy. This structured approach is applicable to a wide range of problems, not just in mathematics but also in everyday life. In conclusion, this problem serves as a valuable exercise in understanding geometric concepts, algebraic equations, and problem-solving strategies. It highlights the interconnectedness of mathematical ideas and their relevance to real-world situations.

Keywords

Rectangle width, perimeter problem, algebraic equation, solve for width, geometric dimensions.