Problem 62 - Elementary Algebra

Question

The perimeter of a rectangle is 50 inches and the area is 126 square inches. Find the dimensions of the rectangle.

Step-by-Step Solution

Answer

The rectangle's dimensions are 18 inches by 7 inches.

Step 1: Understand the Given Information

We are given two pieces of information about the rectangle: its perimeter is 50 inches, and its area is 126 square inches.

Step 2: Define Variables

Let \( l \) represent the length and \( w \) represent the width of the rectangle.

Step 3: Use the Perimeter Formula

The formula for the perimeter of a rectangle is \( P = 2(l + w) \). Given the perimeter is 50 inches, we have the equation: \[ 2(l + w) = 50 \].

Step 4: Simplify the Perimeter Equation

Solve for \( l + w \) by dividing the entire equation by 2:\[ l + w = 25 \].

Step 5: Use the Area Formula

The formula for the area of a rectangle is \( A = lw \). Given the area is 126 square inches, we have the equation:\[ lw = 126 \].

Step 6: Solve the System of Equations

We have two equations:1. \( l + w = 25 \)2. \( lw = 126 \)Use the first equation to express \( l \) in terms of \( w \):\[ l = 25 - w \].

Step 7: Substitute into the Area Equation

Substitute \( l = 25 - w \) into \( lw = 126 \):\[ (25 - w)w = 126 \]This simplifies to:\[ 25w - w^2 = 126 \].

Step 8: Form a Quadratic Equation

Rearrange \( 25w - w^2 = 126 \) into standard quadratic form:\[ w^2 - 25w + 126 = 0 \].

Step 9: Solve the Quadratic Equation

Use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -25 \), and \( c = 126 \).Substitute these values:\[ w = \frac{25 \pm \sqrt{(-25)^2 - 4 \cdot 1 \cdot 126}}{2} \].

Step 10: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \):\[ (-25)^2 - 4 \cdot 126 = 625 - 504 = 121 \].

Step 11: Solve for Width

The equation becomes:\[ w = \frac{25 \pm 11}{2} \]which gives two solutions:1. \( w = \frac{36}{2} = 18 \)2. \( w = \frac{14}{2} = 7 \).

Step 12: Determine the Dimensions

For \( w = 18 \), \( l = 25 - 18 = 7 \).For \( w = 7 \), \( l = 25 - 7 = 18 \).Hence, the dimensions are 18 inches and 7 inches. Both configurations represent the same rectangle.

Key Concepts

System of Equations

A system of equations is a set of equations with multiple variables that we need to solve simultaneously.
In the context of an algebra word problem, like our rectangle dimensions problem, systems of equations often help us relate different pieces of information to find a solution.
For the rectangle, we know:

• The perimeter formula is \( P = 2(l + w) \) which gives us the equation: \( l + w = 25 \).
• The area formula is \( A = lw \) leading to the equation: \( lw = 126 \).

These two equations form a system of equations:

• \( l + w = 25 \)
• \( lw = 126 \)

We solve this system by expressing one variable in terms of the other, then substituting back to solve each equation one by one.
This is a powerful technique because it allows us to handle multiple constraints and find exact values for our variables.

Quadratic Equation

The quadratic equation is a key mathematical concept often appearing in solving systems of equations, particularly in problems involving shapes like rectangles.
A quadratic equation has the general form \( ax^2 + bx + c = 0 \).
In our rectangle problem, after substituting the expression for length \( l \) from one of our system equations, we end up with:

• \( w^2 - 25w + 126 = 0 \)

This equation is quadratic because it involves \( w^2 \), highlighting that solving it involves finding both values of \( w \) that make the equation true.To solve it, we use the quadratic formula:

• \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

For our specific equation, we compute:

• Discriminant: \((-25)^2 - 4 \cdot 1 \cdot 126 = 121\)

The positive discriminant indicates two distinct real solutions, which translates to two possible dimensions for the rectangle.
Finding out these solutions completes the determination of the rectangle's dimensions.

Rectangle Dimensions

Understanding rectangle dimensions involves comprehending how length and width relate to a rectangle's perimeter and area.
In any rectangle:

• Perimeter represents the total distance around the edges, measured by \( P = 2(l + w) \).
• Area is the amount of space inside, calculated by \( A = lw \).

These formulas are crucial because they form the foundation of the equations used in our system.In solving for the dimensions of the rectangle in this algebra word problem, we began by letting \( l \) and \( w \) be the length and width respectively.
Given the perimeter (50 inches) and area (126 square inches), we derived the equations \( l + w = 25 \) and \( lw = 126 \).By expressing one variable in terms of another, such as \( l = 25 - w \), we could substitute and solve using the quadratic method.
Ultimately, this gives us dimensions of either:

• Length \( = 18 \) inches and Width \( = 7 \) inches
• Or vice versa, Length \( = 7 \) inches and Width \( = 18 \) inches.

Both result in the same rectangle, emphasizing that in rectangles, length and width are interchangeable.