Problem 11 - Elementary Algebra Within Reach

Question

In Exercises \(11-16\), use a system of linear equations to find the dimensions of the rectangle that meet the specified conditions. 40 feet The length is 4 feet greater than the width.

Step-by-Step Solution

Answer

The dimensions of the rectangle are: Length = 12 feet, Width = 8 feet.

Step 1: Formulate the Equations

The given information states that the perimeter of the rectangle is 40 feet. The formula for the perimeter of a rectangle is \(2(Length+Width)= Perimeter\). This gives us equation \(1: 2(L+W)=40\). The problem also states that the length is 4 feet greater than the width, giving us equation \(2: L=W+4\). Both the equations combined form a system of linear equations.

Step 2: Substitute Equation 2 into Equation 1

Take equation \(2: L=W+4\) and substitute it into the equation \(1: 2(L+W)=40\). This results in the equation \(2((W + 4) + W) = 40\), which simplifies to \(2(2W + 4) = 40\).

Step 3: Solve for Width

The equation \(2(2W + 4) = 40\) simplifies to \(4W + 8 = 40\). Subtracting 8 from both sides yields \(4W = 32\), then dividing both side by 4 gives \(W = 8\). Hence, the width of the rectangle is 8 feet.

Step 4: Solve for Length

Substitute the width \(W = 8\) into our equation \(2: L = W + 4 = 8 + 4\). So, \(L = 12\). Hence, the length of the rectangle is 12 feet.

Key Concepts

Rectangle Perimeter

Understanding the perimeter of a rectangle is essential in solving many practical problems. In simple terms, the perimeter is the total distance around the edge of the rectangle. For a rectangle, this is found by adding up all the sides: two lengths and two widths. The formula for the perimeter can be represented as:

• The formula: \( P = 2(L + W) \), where \( P \) stands for perimeter, \( L \) for length, and \( W \) for width.
• This formula helps wrap our understanding around the need for both width and length in determining perimeter.

Knowing this, we can tackle practical examples involving rectangles. For instance, if we know the perimeter is 40 feet, we can set up an equation like \( 2(L + W) = 40 \) to find the precise dimensions.

Solving Equations

In this problem, we use a system of linear equations to figure out the dimensions of the rectangle. Solving a system involves finding values for variables that satisfy all the equations at once.

• First, we need to create our equations based on the problem details. We have two key pieces of information: the total perimeter and that the length is 4 feet longer than the width.
• This gives us the following equations: \( 2(L + W) = 40 \) and \( L = W + 4 \).

To find the dimensions, we can substitute one equation into another. Substituting \( L = W + 4 \) into \( 2(L + W) = 40 \), we can solve for one variable at a time.

• After simplifying, first solve for \( W \) (width) and then plug back to find \( L \) (length).

Through these calculated steps, you efficiently determine the measurements of the rectangle.

Geometry Application

Geometry allows us to translate real-life problems into mathematical language. The understanding and application of concepts like perimeter and area can manifest in relatable scenarios, such as designing spaces or crafting objects.

• Consider a scenario where knowing the perimeter helps in determining how much material is needed to decorate the edge of a garden.
• In developing a play area, correctly assessing dimensions ensures safety and functionality.
• Understanding the predictive relationship between dimensions, using equations like \( L = W + 4 \), can guide us in practical applications.

By applying geometric principles and solving related equations, you gain insight into both everyday applications and more complex constructions.