Problem 42 - Precalculus Mathematics for Calculus

Question

A parcel of land is 6 ft longer than it is wide. Each diagonal from one corner to the opposite corner is 174 ft long. What are the dimensions of the parcel?

Step-by-Step Solution

Answer

The parcel's dimensions are 120 ft wide and 126 ft long.

Step 1: Identify Given Variables

We are asked to find the dimensions (length and width) of the parcel of land. Let the width be \( x \) ft. According to the problem, the length is 6 ft longer than the width, so the length is \( x + 6 \) ft. Each diagonal is given as 174 ft.

Step 2: Use the Pythagorean Theorem

Since the land is a rectangle, its diagonal forms the hypotenuse of a right triangle with the length and width as the other two sides. By the Pythagorean theorem: \( x^2 + (x+6)^2 = 174^2 \). Expand and simplify the equation.

Step 3: Solve the Equation

First, expand the equation: \( x^2 + (x+6)^2 = x^2 + 12x + 36 = 174^2 \). Simplify to \( 2x^2 + 12x + 36 = 30276 \). Rearrange to form: \( 2x^2 + 12x - 30240 = 0 \). Divide the entire equation by 2: \( x^2 + 6x - 15120 = 0 \).

Step 4: Apply the Quadratic Formula

The quadratic equation is \( x^2 + 6x - 15120 = 0 \), where \( a = 1 \), \( b = 6 \), and \( c = -15120 \). Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute values: \( x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-15120)}}{2} \).

Step 5: Calculate the Discriminant

Calculate \( b^2 - 4ac = 6^2 - 4\cdot 1\cdot (-15120) = 36 + 60480 = 60516 \). The discriminant is 60516.

Step 6: Find the Roots

Calculate the roots using the quadratic formula: \( x = \frac{-6 \pm \sqrt{60516}}{2} \). The square root of 60516 is approximately 246, so the roots are: \( x = \frac{-6 + 246}{2} \) and \( x = \frac{-6 - 246}{2} \).

Step 7: Determine the Physical Solution

The possible solutions are \( x = 120 \) and \( x = -126 \). Since width cannot be negative, \( x = 120 \) ft is the valid solution. Thus, the width is 120 ft and the length is \( 120 + 6 = 126 \) ft.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). It is called quadratic because "quad" means square, which refers to the \( x^2 \) term.In our problem, we ended up with the quadratic equation \( x^2 + 6x - 15120 = 0 \). Quadratic equations are crucial in problems involving real-world scenarios because they model the relationship between different factors.

• The coefficients \( a \), \( b \), and \( c \) in a quadratic equation determine the shape and position of its graph, a parabola.
• The solutions to the quadratic equation are known as roots, which can be found using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• The term \( b^2 - 4ac \) is called the discriminant, which indicates the nature of the roots: real or complex, and whether they are distinct or repeated.

Understanding how to solve quadratic equations is essential for tackling many algebraic challenges, such as determining the possible dimensions of geometrical shapes like rectangles.

Rectangle Dimensions

Rectangle dimensions refer to the measurements that define the size of a rectangle, specifically the length and width. These dimensions are fundamental in calculating properties like area, perimeter, and diagonal length.In this problem, the width was set as \( x \) and the length as \( x + 6 \). By establishing these variables, we could use algebra to find the exact measurements based on the information provided.

• Width \( x \): The shorter side of the rectangle. In our problem, we assumed this to be \( x \), which helped us to solve the equation involving the diagonal.
• Length \( x + 6 \): Typically, the longer side, influenced by the condition "6 ft longer than the width." This helped establish the correct relationship between the rectangle's sides.

Defining dimensions correctly is crucial to avoid errors in solutions involving geometry, ensuring that each step leads closer to finding the right values.

Problem-Solving Steps

Effective problem-solving involves breaking down a problem into clear, manageable steps, ensuring accurate and consistent results.In this exercise, we followed these essential steps:

• Identify Given Variables: Clearly defining what each variable represents. We chose \( x \) for width and \( x + 6 \) for length.
• Use Known Theorems: The Pythagorean Theorem helped relate the sides of the rectangle to its diagonal.
• Formulate and Simplify the Equation: We created an equation from the information and simplified it to a quadratic form.
• Solve Using Algebraic Methods: Applying the quadratic formula and calculating the roots facilitated finding the physical solution.
• Validate Solutions: Checking the solutions to ensure they meet all given conditions, dismissing any negative length or width.

These steps underscore the importance of a structured approach in problem-solving, preventing missteps and ensuring all parts of the problem are addressed appropriately.

Diagonal of a Rectangle

The diagonal of a rectangle divides it into two right-angled triangles, playing a key role in geometry problems involving rectangles.By the Pythagorean Theorem, the length of the diagonal \( d \) can be expressed as:\[ d = \sqrt{l^2 + w^2} \]where \( l \) is the length and \( w \) is the width of the rectangle.

• In our problem, using the given diagonal length of 174 feet, we established the equation \( x^2 + (x + 6)^2 = 174^2 \).
• The diagonal's length is critical because it links directly to the dimensions through the Pythagorean relationship, allowing for algebraic manipulation of rectangle properties.
• Understanding how to use and solve for the diagonal helps in diverse problems, from real-life applications to purely mathematical inquiries.

Linking diagonal and side properties using geometry ensures that all aspects of a rectangle are accurately quantified, enabling comprehensive solutions to intricate problems.