Problem 36 - Algebra A Combined Function

Question

If the length of a rectangular parking lot is 10 meters less than twice its width, and the perimeter is 400 meters, find the length of the parking lot.

Step-by-Step Solution

Answer

The length of the parking lot is 130 meters.

Step 1: Understand the Relationship

Let's denote the width of the parking lot as \( w \). According to the problem, the length \( l \) is 10 meters less than twice the width. This gives us the equation \( l = 2w - 10 \).

Step 2: Use Perimeter Formula

The perimeter \( P \) of a rectangle is given by \( P = 2l + 2w \). Substituting the given perimeter, we have \( 400 = 2l + 2w \).

Step 3: Substitute for Length

Substitute the expression for \( l \) from Step 1 into the perimeter equation: \( 400 = 2(2w - 10) + 2w \). Simplifying gives us \( 400 = 4w - 20 + 2w \).

Step 4: Solve for Width

Combine like terms in the equation from Step 3: \( 400 = 6w - 20 \). Add 20 to both sides to get \( 420 = 6w \). Divide both sides by 6 to find \( w = 70 \).

Step 5: Find the Length

Now that we have \( w = 70 \), substitute it back into the equation \( l = 2w - 10 \) to find the length: \( l = 2(70) - 10 = 140 - 10 = 130 \).

Key Concepts

Length and Width Relationship

Understanding the relationship between the length and the width of a rectangle can simplify solving problems that involve calculating various measurements, such as a rectangle's perimeter or area. In our original exercise, the length of the parking lot is described in relation to its width. Specifically, the length is 10 meters less than twice the width.
This relationship can be translated into a formula. Let's say the width is represented by the variable \( w \). The length \( l \) is then expressed as \( l = 2w - 10 \).
This equation succinctly captures the relationship: by knowing one side, you can quickly find the other. By grasping this relationship, it becomes simpler to solve for unknowns when part of the measurement is provided or when combined with additional information, like the perimeter.

Solving Linear Equations

Linear equations are fundamental in solving many geometric problems, including those dealing with rectangles. The equation derived from the rectangular perimeter helps us find unknown values given certain conditions.
In our exercise, after relating length to width, we have a linear equation involving both length and width derived from the perimeter formula \( P = 2l + 2w \). With the given perimeter of 400 meters, substituting our expression for the length gives us another equation: \( 400 = 2(2w - 10) + 2w \).
By simplifying further, we solve \( 400 = 4w - 20 + 2w \) into \( 400 = 6w - 20 \). Now, simple algebraic steps allow us to solve for \( w \):

• First, add 20 on both sides: \( 420 = 6w \).
• Then, divide both sides by 6: \( w = 70 \).

This problem demonstrates how linear equations provide a consistent framework for determining unknowns from known values.

Geometric Problem Solving

Geometric problem solving combines mathematical calculations with spatial considerations. When solving problems involving rectangles, it's crucial to effectively use the geometric properties and relationships. The problem we're working with involves both an understanding of rectangles and proficiency in mathematics.
For the rectangle in our scenario, solving the equation based on the perimeter directly ties geometric properties to algebraic manipulation. Specifically, understanding that perimeter involves all sides of the rectangle, the formula \( P = 2l + 2w \) ensures that we account for both the length and width.
Calculating the perimeter allowed us to establish an equation and solve for width, thus enabling us to deduce the length. This kind of geometric problem-solving requires:

• Interpreting problem statements into mathematical equations
• Applying appropriate geometric principles, like the perimeter formula
• Using algebra to solve for unknowns

Once these steps are followed, you can successfully tackle a wide range of geometric problems by leveraging both geometric insights and algebraic methods.