Problem 100 - Algebra and Trigonometry

Question

Dimensions of a Lot A rectangular parcel of land has an area of 5000 \(\mathrm{ft}^{2}\) . A diagonal between opposite comers is measured to be 10 \(\mathrm{ft}\) longer than one side of the parcel. What are the dimensions of the land, rounded to the nearest foot?

Step-by-Step Solution

Answer

The dimensions are approximately 100 ft by 50 ft.

Step 1: Define the Variables

Let the length of the rectangle be \( l \) and the width be \( w \). We are given that the area is 5000 ft², so \( lw = 5000 \).

Step 2: Express Diagonal in Terms of Length

We know the diagonal is 10 ft longer than one side of the parcel. Without loss of generality, assume the diagonal is 10 ft longer than the width. Thus the diagonal \( d = w + 10 \).

Step 3: Apply the Pythagorean Theorem

According to the Pythagorean Theorem, the diagonal of a rectangle is given by \( d = \sqrt{l^2 + w^2} \). Substitute \( d = w + 10 \) into this equation to get \( w + 10 = \sqrt{l^2 + w^2} \).

Step 4: Solve for Length and Width

Substitute \( l = \frac{5000}{w} \) from Step 1 into the equation derived in Step 3: \[ \sqrt{\left(\frac{5000}{w}\right)^2 + w^2} = w + 10 \]Squaring both sides, we get \[ \left(\frac{5000}{w}\right)^2 + w^2 = (w+10)^2 \]Simplify and solve this equation for \( w \).

Step 5: Quadratic Simplification

Expanding both sides gives:\[ \frac{25000000}{w^2} + w^2 = w^2 + 20w + 100 \]This simplifies to:\[ \frac{25000000}{w^2} = 20w + 100 \]Multiplying through by \( w^2 \) provides:\[ 25000000 = 20w^3 + 100w^2 \]Solve for \( w \) using a suitable method such as numerical approximation.

Step 6: Determine Dimensions

Assuming the cubic equation \( 20w^3 + 100w^2 - 25000000 = 0 \) is solved numerically, we find \( w \approx 50 \). Substitute \( w \) into \( l = \frac{5000}{w} \) to get \( l \approx 100 \).

Step 7: Verify Solution

Check that the dimensions satisfy both the area and diagonal condition:- Area \( lw = 5000 \) confirms \( 100 \times 50 = 5000 \).- Diagonal \( \sqrt{100^2 + 50^2} \approx 111.80 \), indeed 10 more than width 50.

Key Concepts

Quadratic Equation

Quadratic equations are algebraic expressions where the highest exponent of the variable is 2. These equations often take the form \( ax^2 + bx + c = 0 \) where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). The key to solving such equations is finding the values of \( x \) that make the equation true, called the roots of the equation.

To solve quadratic equations, we can use a variety of methods:

• Factoring: Express the equation as a product of its factors, if possible, to find the roots.
• Completing the square: This method involves rearranging the equation into a perfect square trinomial.
• Quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) provides a direct solution to the equation.
• Graphing: Plotting the equation to find where the parabola intersects the x-axis.

Quadratic equations often appear in real-world applications like motion predictions, calculating areas, and architectural design. In our problem, solving quadratic expressions helps to determine dimensions that satisfy given conditions like area and diagonal length.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations (such as addition, subtraction, multiplication, and division). They're fundamental in algebra and mathematics because they express relationships in concise form. In the context of solving problems, these expressions help describe equations or inequalities.

When working with algebraic expressions, several key steps include:

• Simplification: Combine like terms and perform operations within expressions for easier manipulation.
• Substitution: Replace variables with known values to solve or verify expressions.
• Expansion: Distribute factors to remove parentheses and simplify further.
• Factorization: Express the expression as a product of its factors when needed.

Understanding how to manipulate these expressions is crucial for solving various mathematical problems. In our original exercise, substituting and rearranging algebraic expressions helps form the equation solved by the quadratic method.

Rectangular Area Calculation

Rectangular area calculation is fundamental in geometry, helping us determine the size of two-dimensional spaces. A rectangle's area is calculated using the formula \( Area = length \times width \). This simple yet essential concept is used in diverse fields from construction to landscaping.

Key considerations in rectangular area problems are:

• Dimensions: The length and width are essential components for calculating area.
• Units: Ensure consistent measurement units to obtain accurate area calculations.
• Applications: Used in real-world scenarios, such as determining land sizes, flooring requirements, and space planning.

In the provided problem, knowing the area helps derive necessary equations connecting width and length. Therefore, understanding the relationship between these dimensions alongside the area formula is crucial for finding specific values that solve a given condition such as additional constraints on diagonals.