Problem 22 - Elementary Algebra

Question

The width of a rectangle is 1 foot more than one-third of its length. If the perimeter of the rectangle is 74 feet, find the area of the rectangle.

Step-by-Step Solution

Answer

The area of the rectangle is 270 square feet.

Step 1: Identify Variables

Let's assign variables to the dimensions of the rectangle. Let \( L \) be the length of the rectangle (in feet), and \( W \) be the width. According to the problem, the width is 1 foot more than a third of the length. We can express this as \( W = \frac{1}{3}L + 1 \).

Step 2: Use Perimeter Formula

The perimeter \( P \) of a rectangle is calculated as \( P = 2L + 2W \). We know the perimeter is 74 feet, so we substitute to get the equation: \( 2L + 2W = 74 \).

Step 3: Substitute Width Expression

Substitute the expression for \( W \) from Step 1 into the perimeter equation: \( 2L + 2\left(\frac{1}{3}L + 1\right) = 74 \).

Step 4: Simplify the Equation

Distribute and combine like terms in the equation: \( 2L + \frac{2}{3}L + 2 = 74 \). Simplifying, we have \( \frac{8}{3}L + 2 = 74 \).

Step 5: Solve for Length

Subtract 2 from both sides to isolate terms with \( L \): \( \frac{8}{3}L = 72 \). Now multiply both sides by \( \frac{3}{8} \) to solve for \( L \): \( L = 27 \) feet.

Step 6: Calculate the Width

Substitute \( L = 27 \) back into the expression for width: \( W = \frac{1}{3}\times 27 + 1 \). Simplifying gives \( W = 9 + 1 = 10 \) feet.

Step 7: Find the Area

The area \( A \) of the rectangle is given by \( A = L \times W \). Substitute the known values: \( A = 27 \times 10 = 270 \).

Step 8: Verify Solution

Check that these dimensions satisfy the conditions of the problem: the perimeter is \( 2\times 27 + 2\times 10 = 54 + 20 = 74 \), confirming our solution is correct.

Key Concepts

Perimeter of a Rectangle

The perimeter is the total distance around the outside of the rectangle. Calculating it helps us understand the relationship between the rectangle's sides. To find the perimeter, you can use the formula:\[ P = 2(L + W) \]where:

• \( L \) is the length of the rectangle
• \( W \) is the width of the rectangle
• \( P \) stands for perimeter

In our exercise, replacing with known values like the fact that the perimeter is 74 feet aids in setting up an equation that can be solved to find the dimensions of the rectangle. Understanding that the perimeter is twice the sum of the length and width helps in isolating these unknowns when given a numerical perimeter.

Length and Width of Rectangle

Understanding the length and width relations in a rectangle is key to solving this type of problem. Here, the width is defined as being 1 foot more than a third of the length, represented algebraically as:\[ W = \frac{1}{3}L + 1 \]This equation allows us to relate the two dimensions. To solve for both length and width when given the perimeter, substitution can be used to plug this width expression into the perimeter equation. For example:

• Substituting \( W = \frac{1}{3}L + 1 \) into the perimeter equation \( P = 2L + 2W \) transforms our problem into one solvable algebraic equation.
• With this setup, we can then simplify further and solve to find the values of \( L \) and \( W \).

This approach neatly handles the dependency of one variable on another and highlights how algebra helps in problem-solving strategies.

Algebraic Equations

Algebraic equations are the building blocks for solving many mathematical problems, including our rectangle exercise. By assigning variables to known and unknown quantities, these equations allow us to form relationships and solve for values we need.In our case, the exercise gave us:

• A relationship for width: \( W = \frac{1}{3}L + 1 \)
• The perimeter equation: \( 2L + 2W = 74 \)

Substituting the width expression into the perimeter equation is a classic use of substitution in algebra, providing an equation we can simplify and solve. Combining terms help isolate the desired variable, which in our example led to knowing that the length \( L = 27 \) feet by solving:\[ \frac{8}{3}L + 2 = 74 \]Algebra not only gives us these equations but guides us through simplification, substitution, and ultimately finding the numerical solution for both length and width.