Finding Perimeter When Length Is Twice The Width And Area Is 128 Yd^2

Introduction

In the realm of geometry, rectangles hold a fundamental place, and understanding their properties is crucial for various applications. Among these properties, the relationship between a rectangle's length, width, area, and perimeter is particularly important. This article delves into a specific problem involving a rectangle where the length is twice its width, and the area is given as 128 square yards. Our primary goal is to determine the perimeter of this rectangle. By carefully applying geometric principles and algebraic techniques, we will systematically solve this problem, providing a clear and concise explanation of each step.

Understanding the Fundamentals of Rectangles

To effectively tackle this problem, let's first establish a solid understanding of the fundamental properties of rectangles. A rectangle is a quadrilateral with four right angles. Its defining characteristics are its length (l), which represents the longer side, and its width (w), which represents the shorter side. The area (A) of a rectangle, which signifies the space it occupies, is calculated by multiplying its length and width: A = l * w. The perimeter (P) of a rectangle, which represents the total length of its boundary, is calculated by adding up the lengths of all its sides: P = 2l + 2w. These basic formulas serve as the foundation for solving problems involving rectangles, including the one at hand.

Setting up the Equations

Now that we have a firm grasp of the basic properties of rectangles, let's translate the problem's given information into mathematical equations. The problem states that the length of the rectangle is twice its width, which we can express as l = 2w. We are also given that the area of the rectangle is 128 square yards, which can be written as A = 128. Using the formula for the area of a rectangle, we can substitute the given values and obtain the equation l * w = 128. This equation, along with the relationship between length and width (l = 2w), forms a system of equations that we can solve to find the dimensions of the rectangle.

Solving for Width and Length

To solve for the width and length of the rectangle, we can use the method of substitution. Since we know that l = 2w, we can substitute this expression for l into the area equation: (2w) * w = 128. Simplifying this equation, we get 2w^2 = 128. Dividing both sides by 2, we have w^2 = 64. Taking the square root of both sides, we find that w = 8 yards. Now that we have the width, we can use the relationship l = 2w to find the length: l = 2 * 8 = 16 yards. Therefore, the width of the rectangle is 8 yards, and the length is 16 yards.

Calculating the Perimeter

With the width and length of the rectangle determined, we can now calculate its perimeter. Using the formula for the perimeter of a rectangle, P = 2l + 2w, we substitute the values we found: P = 2 * 16 + 2 * 8. Simplifying this expression, we get P = 32 + 16 = 48 yards. Therefore, the perimeter of the rectangle is 48 yards. This completes the solution to the problem, as we have successfully determined the perimeter of the rectangle given its area and the relationship between its length and width.

Alternative approach to find perimeter

Introduction to Perimeter Calculation Methods

In the previous section, we meticulously calculated the perimeter of the rectangle by first determining its width and length. However, there exists an alternative approach that allows us to find the perimeter directly, without explicitly calculating the individual dimensions. This method leverages the relationship between the area, length-to-width ratio, and perimeter of a rectangle. By understanding and applying this alternative approach, we gain a deeper appreciation for the interconnectedness of geometric properties and expand our problem-solving toolkit.

Deriving the Alternative Formula

To derive the alternative formula for calculating the perimeter, we begin with the fundamental formulas for the area and perimeter of a rectangle: A = l * w and P = 2l + 2w. We are given that the length is twice the width, which we can express as l = 2w. Our goal is to express the perimeter solely in terms of the area. Substituting l = 2w into the area equation, we get A = (2w) * w = 2w^2. Solving for w^2, we have w^2 = A / 2. Taking the square root of both sides, we find w = √(A / 2). Now, substituting l = 2w into the perimeter equation, we get P = 2(2w) + 2w = 6w. Finally, substituting w = √(A / 2) into the perimeter equation, we obtain the alternative formula: P = 6√(A / 2). This formula allows us to calculate the perimeter directly from the area and the length-to-width ratio.

Applying the Alternative Formula

Now that we have derived the alternative formula, let's apply it to the problem at hand. We are given that the area of the rectangle is 128 square yards. Substituting A = 128 into the formula P = 6√(A / 2), we get P = 6√(128 / 2) = 6√64. Since the square root of 64 is 8, we have P = 6 * 8 = 48 yards. This result matches the perimeter we calculated using the previous method, confirming the validity of the alternative formula. This alternative approach provides a more direct route to the solution, highlighting the power of algebraic manipulation and formula derivation in problem-solving.

Comparing the Two Approaches

Both approaches, the traditional method of first finding the width and length and the alternative formula-based method, lead to the same correct answer. However, they offer different perspectives on the problem and employ different problem-solving strategies. The traditional method relies on a step-by-step approach, systematically solving for the unknown dimensions before calculating the perimeter. This method is intuitive and easy to follow, making it suitable for those who prefer a more structured approach. The alternative method, on the other hand, utilizes algebraic manipulation to derive a direct formula for the perimeter. This method is more concise and efficient, but it requires a deeper understanding of the relationships between geometric properties. The choice between the two approaches depends on individual preferences and problem-solving styles. Some may find the traditional method more straightforward, while others may appreciate the elegance and efficiency of the alternative formula.

Conclusion

In this article, we tackled the problem of finding the perimeter of a rectangle given its area and the relationship between its length and width. We explored two distinct approaches: the traditional method of first determining the width and length and the alternative method using a derived formula. Both methods yielded the same result, demonstrating the versatility of geometric problem-solving. By understanding the fundamental properties of rectangles and mastering algebraic techniques, we can confidently solve a wide range of geometric problems. The key takeaway is that a flexible and adaptable approach, combined with a solid foundation in geometric principles, is essential for success in mathematical problem-solving.