Problem 53 - Intermediate Algebra

Question

Set up an equation and solve each problem. The perimeter of a rectangle is 44 inches, and its area is 112 square inches. Find the length and width of the rectangle.

Step-by-Step Solution

Answer

The length is 14 inches and the width is 8 inches.

Step 1: Understand the Problem

We are given the perimeter and area of a rectangle. The perimeter is 44 inches, and the area is 112 square inches. We need to find the length and width of this rectangle.

Step 2: Recall Formulas

The formula for the perimeter of a rectangle is: \(P = 2 \times (l + w)\), where \(l\) is the length and \(w\) is the width. The formula for the area of a rectangle is: \(A = l \times w\).

Step 3: Set Up Equations

Using the formulas, write the equations from the information given.1. For perimeter: \(2 \times (l + w) = 44\)2. For area: \(l \times w = 112\)

Step 4: Simplify Perimeter Equation

Simplify the perimeter equation for easier calculation: \(l + w = 22\)

Step 5: Express One Variable in Terms of Another

Solve the equation \(l + w = 22\) for one of the variables. For example, express \(l\) in terms of \(w\): \(l = 22 - w\)

Step 6: Substitute in Area Equation

Substitute \(l = 22 - w\) into the area equation: \((22 - w) \times w = 112\)

Step 7: Solve Quadratic Equation

Expand and rearrange \(22w - w^2 = 112\) to form a quadratic equation \(w^2 - 22w + 112 = 0\). Solve this for \(w\) using the quadratic formula: \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -22\), and \(c = 112\).

Step 8: Calculate Discriminant and Solve

Calculate the discriminant \(D = b^2 - 4ac = 484 - 448 = 36\). Since the discriminant is positive, there are two solutions:\(w = \frac{22 \pm 6}{2}\).Therefore, the solutions for \(w\) are \(w = 14\) and \(w = 8\).

Step 9: Find Corresponding Lengths

For \(w = 14\), substitute back to find \(l = 22 - 14 = 8\). For \(w = 8\), substitute back to find \(l = 22 - 8 = 14\).

Step 10: Conclusion

Both pairs \((l, w) = (14, 8)\) and \((l, w) = (8, 14)\) satisfy the conditions of the perimeter and area. Thus, the length is 14 inches and the width is 8 inches, or vice versa.

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle is the total length around the rectangle. It consists of the lengths of all four sides added together. If you think of taking a walk around the edge of a rectangle, the perimeter tells you the total distance you'd travel.

To find the perimeter of a rectangle, use this straightforward formula:

• Perimeter (P) = 2 times the sum of the length (l) and the width (w): \( P = 2 \times (l + w) \).

Using this formula, you can calculate the perimeter if you know the length and width. Remember, the length and width are simply the horizontal and vertical spans of the rectangle.

In this exercise, you know that the perimeter is 44 inches. By substituting the perimeter value into the formula, you set up the equation \( 2 \times (l + w) = 44 \). Simplifying gives \( l + w = 22 \), preparing you to solve for each dimension of the rectangle.

Area of a Rectangle

The area of a rectangle measures how much space is contained within its four sides. It's the amount of surface covered by the rectangle, which you might imagine as the number of square tiles needed to completely fill an empty rectangle.

To determine the area, you multiply the length by the width:

• Area (A) = length (l) times width (w): \( A = l \times w \).

Knowing both the length and width allows you to easily compute the area.

In the given problem, the area is 112 square inches. This means when you multiply the length by the width, the result equals 112. By combining this information with the perimeter equation, you can solve for both length and width to discover the sizes that fit both conditions.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2, and they often look like this: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.

In rectangle problems, they can arise when you have two variables closely connected, such as in our exercise. After expressing length in terms of width or vice versa, you may substitute one variable from a linear equation into an area equation, resulting in a quadratic equation.

For the exercise at hand, after rewriting the equation with width expressions, you form \( w^2 - 22w + 112 = 0 \). Solving this requires techniques such as factoring or the quadratic formula:

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

Calculating the discriminant \( D = b^2 - 4ac \), helps us determine the nature of the roots. A positive \( D \) indicates two distinct real solutions. Simplifying the quadratic equation gives the width values that satisfy both perimeter and area conditions of the rectangle.