Problem 275 - Elementary Algebra

Question

In the following exercises, solve using rectangle properties. A rectangular parking lot has perimeter 250 feet. The length is five feet more than twice the width. Find the length and width of the parking lot.

Step-by-Step Solution

Answer

The width is 40 feet, and the length is 85 feet.

Step 1: Define Variables

Let the width of the parking lot be represented by the variable \(w\). According to the problem, the length is five feet more than twice the width. Therefore, let the length be represented by \(l = 2w + 5\).

Step 2: Write the Perimeter Formula for a Rectangle

The formula for the perimeter of a rectangle is given by \(P = 2l + 2w\). Given that the perimeter is 250 feet, we can write the equation \(250 = 2l + 2w\).

Step 3: Substitute the Expression for Length

Substitute the expression \(l = 2w + 5\) into the perimeter equation: \(250 = 2(2w + 5) + 2w\).

Step 4: Simplify the Equation

Distribute and combine like terms in the equation: \(250 = 4w + 10 + 2w\). Simplify it further to \(250 = 6w + 10\).

Step 5: Solve for the Width

Isolate \(w\) by subtracting 10 from both sides: \(240 = 6w\). Then divide both sides by 6 to find the width: \(w = 40\).

Step 6: Calculate the Length

Now that the width is known, substitute \(w = 40\) back into the expression for length: \(l = 2(40) + 5\). This gives \(l = 80 + 5 = 85\).

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle is the total distance around its edges.
You can find it by adding up the lengths of all four sides.
The formula for perimeter (\(P\)) is simple:
\(P = 2l + 2w\)
Where \(l\) stands for length and \(w\) stands for width.
In this problem, we know the perimeter is 250 feet. This key information helps set up our equation for solving the problem.
To solve the problem, we will manipulate this formula to find the unknowns: length and width.

Algebraic Equations

Algebraic equations are mathematical statements that show the equality between two expressions.
In problems like our parking lot example, equations are used to set relations between variables and constants.

Our problem uses the formula for perimeter, which gives us: \(250 = 2l + 2w\).
This equation is a stepping stone. We feed in more information to find solutions for the width and length.
Turning word problems into algebraic equations makes them easier to solve. The goal is to isolate variables, like replacing 'length' with '2w + 5', which simplifies our calculations.

Geometry Problem-Solving

Geometry is all about shapes and solving for their properties.
Here, we deal with a rectangle, a very common shape. Understanding the properties and formulas helps in solving equations.

Our problem uses the perimeter property of a rectangle to establish relations. We then break down these relations into simpler algebraic steps.
By substituting and simplifying, complex geometry problems turn into easier algebra problems.
This method applies to various geometric figures, making problem-solving structured and less daunting.

Variable Substitution

Variable substitution means replacing one variable with a known expression.
In our exercise, we let the width be \(w\) and express the length as \(l=2w+5\).
This substitution helps us rewrite the perimeter formula into a single-variable equation.

Substituting \(l\) into \(2l + 2w = 250\) gives: \(2(2w + 5) + 2w = 250\).
This step simplifies our calculations by focusing on one variable at a time.
From there, solving the equation \(250 = 6w +10\) becomes straightforward. This approach streamlines solving similar algebra and geometry problems.