Problem 41 - College Algebra

Question

Find the dimensions of an American football fi ld. The length is \(200 \mathrm{ft}\) ore than the width, and the perimeter is \(1,040 \mathrm{ft}\). Fi \(\mathrm{d}\) the length and width. Use the perimeter formula \(P=2 L+2 W\).

Step-by-Step Solution

Answer

The width of the field is 160 ft and the length is 360 ft.

Step 1: Understand the problem

We start by defining our variables. Let \( L \) be the length of the field and \( W \) be the width. We are given two pieces of information: \( L = 200 + W \) and \( P = 1040 \) ft, where the perimeter formula for a rectangle is \( P = 2L + 2W \).

Step 2: Substitute the length in the perimeter equation

Using the information \( L = 200 + W \), substitute \( L \) in the perimeter equation. This gives us:\[P = 2(200 + W) + 2W = 1040\]

Step 3: Simplify the equation

Distribute the 2 in the equation:\[2 \times 200 + 2W + 2W = 1040\]Which simplifies to:\[400 + 4W = 1040\]

Step 4: Solve for the width

Rearrange the equation to isolate \( W \):\[4W = 1040 - 400\]\[4W = 640\]Divide both sides by 4:\[W = 160\]

Step 5: Calculate the length

Now that we have \( W = 160 \), use the equation \( L = 200 + W \) to find \( L \):\[L = 200 + 160 = 360\]

Step 6: Verify the solution

Check the calculated dimensions by ensuring the perimeter sums correctly:\[P = 2L + 2W = 2(360) + 2(160) = 720 + 320 = 1040\]The sum matches the given perimeter, confirming the solution.

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle is essentially the total distance around the edges of the rectangle. For a rectangle, the formula used to calculate the perimeter is given by:

• \( P = 2L + 2W \)

where \( L \) is the length and \( W \) is the width of the rectangle.
In our problem, the total perimeter of the football field is provided as \( 1040 \) ft. To find the missing dimensions, we use this formula alongside other given information. The perimeter helps us establish a relationship between the length and the width, which can then be used to solve for the unknown dimensions using algebra.
Understanding the concept of perimeter is crucial as it forms the fundamental basis for many geometric calculations.

Rectangular Dimensions

When dealing with rectangular dimensions, knowing both the length and width will allow you to calculate the area and perimeter of the rectangle, which are commonly sought values in geometry problems.
The key information in this problem is that the length of the rectangle is \( 200 \) feet more than the width. This can be expressed in mathematical terms as:

• \( L = W + 200 \)

This relationship is crucial for substituting in the perimeter formula and helps solve the problem effectively. Exploiting such relationships simplifies the calculations and points us in the direction of using algebraic manipulation to find the desired lengths.

Substitution Method

The substitution method is a popular algebraic technique used to find unknown variables, especially in problems involving equations. Here is a simple breakdown of using substitution:

• Identify variable relationships, like \( L = 200 + W \).
• Replace one variable in an equation with its equivalent expression in terms of another variable.

In this exercise, once we have confirmed that \( L = 200 + W \), we replace \( L \) in the perimeter equation:

• \( P = 2(200 + W) + 2W = 1040 \)

This simplifies the equation to involve only one variable \( W \), which can then be solved using standard algebraic techniques. The substitution method is particularly efficient in scenarios where relationships between variables are already established.

Algebraic Expressions

An algebraic expression represents a mathematical relationship using numbers, variables, and arithmetic operations. In our scenario, we experience expressions forming from given problem statements, such as:

• \( L = 200 + W \)
• \( P = 2L + 2W \)

To solve for \( W \) and \( L \), simplifying and manipulating these expressions is essential. By distributing values and combining like terms, the problem statement is converted into a solvable equation. Algebraic expressions are powerful tools to translate written problems into mathematical ones and then solve them through systematic methods.
For instance, we rearrange and simplify the perimeter equation to find \( W \) quickly through subtraction and division, illustrating the practical utility of algebra in everyday problem-solving.