Rectangle Dimensions Unveiled Length Is 2 More Than Three Times Width
In the realm of geometry, rectangles hold a fundamental position, their symmetrical structure and predictable properties making them essential building blocks for more complex shapes and calculations. This article delves into a fascinating problem involving a rectangle, where the relationship between its length and width is defined by a specific equation, and its area is known. Our mission is to unravel the dimensions of this rectangle, employing algebraic techniques and problem-solving strategies to arrive at the solution. This exploration will not only enhance our understanding of rectangles but also sharpen our mathematical prowess.
Decoding the Rectangle's Dimensions: A Step-by-Step Approach
Let's embark on a journey to decipher the dimensions of this enigmatic rectangle. The heart of the problem lies in the given information: the length is intricately linked to the width, described as "2 more than three times the width." This verbal expression can be translated into a concise algebraic equation, a cornerstone of our solution. The area, a crucial piece of the puzzle, is provided as 161 square inches. Our goal is to weave these clues together, using the power of algebra to unveil the rectangle's true nature.
Defining the Variables: The Language of Algebra
In the world of algebra, variables serve as placeholders for unknown quantities, allowing us to manipulate and solve equations. Let's introduce the variable x to represent the width of the rectangle. This seemingly simple step unlocks a chain of deductions. Since the length is defined in terms of the width, we can express the length using the same variable. According to the problem statement, the length is "2 more than three times the width." Translating this into algebraic language, we get the expression 3_x_ + 2 for the length.
With the width represented by x and the length by 3_x_ + 2, we've successfully captured the essence of the rectangle's dimensions in algebraic terms. This is a crucial step, as it allows us to formulate an equation that connects these dimensions to the known area.
Area of a Rectangle: The Key Equation
The area of a rectangle is a fundamental concept, defined as the product of its length and width. In mathematical notation, this translates to: Area = Length × Width. This seemingly simple equation is the key to unlocking the rectangle's dimensions. We know the area is 161 square inches, and we've expressed the length and width in terms of x. Substituting these values into the area equation, we get:
161 = (3_x_ + 2) × x
This equation encapsulates the relationship between the rectangle's dimensions and its area. It's a quadratic equation, a type of equation that involves a variable raised to the power of 2. Solving this equation will reveal the value of x, which in turn will allow us to determine the width and length of the rectangle.
Solving the Quadratic Equation: A Journey Through Algebra
To solve the quadratic equation 161 = (3_x_ + 2) × x, we need to rearrange it into a standard form. This involves expanding the equation and moving all terms to one side, resulting in an equation of the form a_x_² + b_x_ + c = 0. Let's break down the steps:
- Expand the equation: 161 = 3_x_² + 2_x_
- Move all terms to one side: 0 = 3_x_² + 2_x_ - 161
Now we have a quadratic equation in standard form. There are several methods for solving quadratic equations, including factoring, completing the square, and the quadratic formula. In this case, factoring might be a bit challenging, so let's employ the quadratic formula, a reliable method that always yields the solutions.
The Quadratic Formula: A Universal Solver
The quadratic formula is a powerful tool that provides the solutions to any quadratic equation in the form a_x_² + b_x_ + c = 0. The formula is:
x = (-b ± √(b² - 4ac)) / (2a)
In our equation, 3_x_² + 2_x_ - 161 = 0, we have a = 3, b = 2, and c = -161. Plugging these values into the quadratic formula, we get:
x = (-2 ± √(2² - 4 × 3 × (-161))) / (2 × 3)
Simplifying this expression, we get two possible values for x:
x = (-2 ± √(1936)) / 6
x = (-2 ± 44) / 6
This gives us two solutions:
_x_₁ = (-2 + 44) / 6 = 7
_x_₂ = (-2 - 44) / 6 = -7.67
Since the width of a rectangle cannot be negative, we discard the negative solution. Therefore, the width of the rectangle is x = 7 inches.
Calculating the Length: Completing the Puzzle
Now that we've found the width, we can easily calculate the length using the expression we derived earlier: Length = 3_x_ + 2. Substituting x = 7, we get:
Length = 3 × 7 + 2 = 23 inches
Thus, the dimensions of the rectangle are width = 7 inches and length = 23 inches.
Verifying the Solution: Ensuring Accuracy
To ensure the accuracy of our solution, we can plug the values of length and width back into the area equation: Area = Length × Width. Substituting the values we found, we get:
Area = 23 inches × 7 inches = 161 square inches
This matches the given area, confirming that our solution is correct. We have successfully unveiled the dimensions of the rectangle!
Formulating the Equation: A Crucial Step in Problem Solving
In the process of solving this problem, a crucial step is formulating the equation that represents the given information. The problem states: "The length of a rectangle is 2 more than three times the width. The area of the rectangle is 161 square inches." If x represents the width of the rectangle, the equation that captures this relationship and is used in the solution process is:
(3_x_ + 2) * x = 161
This equation embodies the essence of the problem, connecting the length, width, and area in a single algebraic expression. It serves as the foundation for solving the problem and finding the dimensions of the rectangle.
Conclusion: A Triumph of Algebra and Geometry
We've successfully navigated the intricacies of this rectangle problem, employing algebraic techniques and problem-solving strategies to unveil its dimensions. By translating the verbal description into algebraic expressions, formulating the area equation, and solving the quadratic equation, we've demonstrated the power of mathematics in unraveling geometric mysteries. This journey has not only provided us with the dimensions of a specific rectangle but has also reinforced our understanding of algebraic concepts and problem-solving skills. The world of mathematics is filled with such fascinating puzzles, waiting to be explored and conquered.
This exploration highlights the importance of translating word problems into mathematical equations. By carefully defining variables, expressing relationships algebraically, and applying appropriate formulas, we can unlock solutions to seemingly complex problems. The rectangle problem serves as a testament to the elegance and power of mathematical reasoning.
