Problem 38 - Algebra and Trigonometry

Question

Dimensions of a Lot A half-acre building lot is five times as long as it is wide. What are its dimensions? [Note: 1 acre \(=43,560 \mathrm{ft}^{2}\) .

Step-by-Step Solution

Answer

The lot is 66 ft wide and 330 ft long.

Step 1: Determine the total area in square feet

A half-acre lot has an area of \( \frac{1}{2} \times 43,560 \text{ ft}^2 = 21,780 \text{ ft}^2 \). This is the total area of the lot.

Step 2: Set up the relationship between length and width

According to the problem, the length (\( L \)) is five times the width (\( W \)). Therefore, we can express this relationship as \( L = 5W \).

Step 3: Express area in terms of width

The area of the lot can be expressed as the product of its length and width: \( L \times W = 21,780 \text{ ft}^2 \). Substitute the expression from Step 2 into this equation to get \( 5W \times W = 21,780 \text{ ft}^2 \), or \( 5W^2 = 21,780 \).

Step 4: Solve for width

To find \( W \), first, divide both sides by 5: \( W^2 = \frac{21,780}{5} = 4,356 \). Then take the square root of both sides: \( W = \sqrt{4,356} \approx 66 \text{ ft} \).

Step 5: Solve for length

Substitute the value of \( W \) into the equation from Step 2 to find the length: \( L = 5 \times 66 = 330 \text{ ft} \). This gives us the dimensions of the lot.

Key Concepts

Area Calculation

Understanding how to calculate the area of land is essential for solving algebra word problems involving dimensions. When dealing with areas, especially in real estate or similar contexts, we often convert acres into square feet for ease of calculation.
In this exercise, we deal with a half-acre lot. To find the total area in square feet, multiply the number of acres by the conversion factor. An acre is equivalent to 43,560 square feet. Thus, a half-acre is calculated as:

• Half-acre = \(\frac{1}{2} \times 43,560 \text{ ft}^2 = 21,780 \text{ ft}^2\)

This gives us the total area upon which further calculations are based. Always ensure to know the conversion, as it plays a pivotal role in problem-solving.

Length and Width Relationship

Understanding relationships between dimensions such as length and width is crucial. Here, the problem states that the length of the lot is five times its width, which is expressed as a simple algebraic equation: \(L = 5W\).
This relationship is vital because it allows you to substitute one variable with the other in an equation. By expressing the length in terms of the width, it simplifies calculating other unknowns.
The substitution not only helps in solving the equations but also in understanding how proportional relationships can guide the calculation. Always look for such relationships to lay a clearer path to reaching the solution more efficiently.

Solving Equations

The final step in this problem is solving mathematical equations to find the specific dimensions. After framing the relationship between the length and width, you express the area as \(L \times W = 21,780 \text{ ft}^2\).
Substitute \(L\) from the relationship to get: \(5W \times W = 21,780\). Simplifying further, you have: \(5W^2 = 21,780\).

• Divide both sides by 5: \(W^2 = 4,356\)
• Take the square root: \(W = \sqrt{4,356} \approx 66 \text{ ft}\)

Once you calculate the width, plug it back into the length equation \(L = 5W\) to get the length: \(L = 330 \text{ ft}\).
Knowing how to manipulate equations helps simplify and solve complex algebraic problems, refining your problem-solving skills.