Problem 28 - Algebra and Trigonometry

Question

Width of a Pasture A pasture is twice as long as it is wide. Its area is \(115,200 \mathrm{ft}^{2}\) . How wide is the pasture?

Step-by-Step Solution

Answer

The width of the pasture is 240 feet.

Step 1: Identify variables

Let the width of the pasture be represented by \( w \). Since the pasture is twice as long as it is wide, the length can be represented as \( 2w \).

Step 2: Write the area formula

The area \( A \) of a rectangle is given by the formula \( A = ext{length} \times ext{width} \). So, the equation for this pasture becomes \( 115,200 = 2w \times w \).

Step 3: Simplify the equation

We start with the equation \( 115,200 = 2w^2 \). Divide both sides by 2 to get \( w^2 = 57,600 \).

Step 4: Solve for \( w \)

Find the square root of both sides of the equation \( w^2 = 57,600 \) to solve for \( w \). So, \( w = \sqrt{57,600} \).

Step 5: Calculate the width

Calculate \( \sqrt{57,600} \) which gives \( w = 240 \). Hence, the width of the pasture is 240 feet.

Key Concepts

Understanding Pasture Dimensions

When we talk about the dimensions of a pasture, especially in a problem related to geometry and algebra, we usually define two key measurements: length and width. In this exercise, the length and width are linked by a specific ratio. Here, the pasture is described as being "twice as long as it is wide." This means if the width is represented by the variable \( w \), then the length can be expressed as \( 2w \).
Knowing how to set up these relationships is crucial because it allows us to use algebraic equations to find missing values. By understanding how dimensions relate, we can simplify complex problems into manageable equations. This approach is not just useful here but is a fundamental skill in solving many real-world problems involving dimensions.

Solving Quadratic Equations

To find the width of the pasture, the problem simplifies into solving a quadratic equation. This happens because when you multiply the length \( 2w \) by the width \( w \), it results in an equation involving \( w^2 \).
Given the area is \( 115,200 \) square feet, we use the area formula for rectangles: \( A = \text{length} \times \text{width} \). Substituting in our values, we end up with:

• \( 115,200 = 2w^2 \)
• Dividing both sides by 2 gives us \( w^2 = 57,600 \)

Quadratic equations typically take the form of \( ax^2 + bx + c = 0 \). Here we solve a simplified version, where only the square term \( w^2 \) is present. Finding the square root of both sides allows us to solve for \( w \). This process is an example of solving by isolation, which is often used in algebra to find unknowns.

Exploring the Length-Width Relationship

The length-width relationship plays a pivotal role in many geometric problems. In our pasture scenario, this relationship helps us set up the correct equation right from the start.
Here, the relationship is simple: the length is double the width. By establishing that the length is \( 2w \) when the width is \( w \), we can substitute these expressions into the area formula.
This substitution is what allowed us to create the equation \( 115,200 = 2w^2 \). Without understanding this intrinsic relationship, setting up the problem correctly would be difficult. This concept highlights the importance of carefully analyzing and interpreting word problems to translate descriptions into mathematical models effectively.