Problem 68 - Intermediate Algebra

Question

Geometry. The area of a rectangle is 63 square centimeters, and its perimeter is 32 centimeters. Find the dimensions of the rectangle.

Step-by-Step Solution

Answer

The rectangle's dimensions are 7 cm and 9 cm.

Step 1: Define Variables for Dimensions

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

Step 2: Use Area Formula

The formula for the area of a rectangle is given by \( l \times w = 63 \). This equation helps us establish a relationship between length and width.

Step 3: Use Perimeter Formula

The formula for the perimeter of a rectangle is given by \( 2(l + w) = 32 \). Simplifying this, we get \( l + w = 16 \).

Step 4: Express Length in Terms of Width

From \( l + w = 16 \), express \( l \) as \( l = 16 - w \).

Step 5: Substitute and Solve for Width

Substitute \( l = 16 - w \) into the area equation: \((16 - w)w = 63\). Expanding gives \( 16w - w^2 = 63 \), which simplifies to \( w^2 - 16w + 63 = 0 \).

Step 6: Solve the Quadratic Equation

Use the quadratic formula: \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -16 \), \( c = 63 \). Calculate the discriminant: \( (-16)^2 - 4 \cdot 1 \cdot 63 = 256 - 252 = 4 \). So, \( w = \frac{16 \pm \sqrt{4}}{2} \).

Step 7: Calculate Possible Widths

Simplify to find \( w = \frac{16 \pm 2}{2} \), giving solutions \( w = 9 \) and \( w = 7 \).

Step 8: Determine Corresponding Lengths

For \( w = 7 \), \( l = 16 - 7 = 9 \). For \( w = 9 \), \( l = 16 - 9 = 7 \).

Step 9: Conclude Dimensions

The rectangle's dimensions are \( 7 \text{ cm} \times 9 \text{ cm} \) or \( 9 \text{ cm} \times 7 \text{ cm} \), both satisfy the conditions given.

Key Concepts

Quadratic Equation

When solving problems involving dimensions of a rectangle, especially when both area and perimeter are involved, quadratic equations often emerge. A quadratic equation is a second-degree polynomial equation of the form \( ax^2 + bx + c = 0 \). In this context, the variables \( l \) and \( w \) represent the length and width of the rectangle, respectively. \( a \), \( b \), and \( c \) are numerical coefficients.Quadratic equations are crucial because they help to compute unknowns when a relationship is represented via squares or products. For instance, in the problem, the relationship \( w^2 - 16w + 63 = 0 \) emerges while linking the area formula and the perimeter formula. Solving this equation involves:

• Calculating the discriminant: \( b^2 - 4ac \) which determines the nature of the roots.
• Applying the quadratic formula: \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Once solved, the quadratic formula yields the potential dimensions of the rectangle, providing either one or two possible solutions for \( w \) and consequently for \( l \). This methodological approach allows us to handle complex relations in geometric figures with precision.

Perimeter and Area

Understanding the concepts of perimeter and area is key to solving rectangle problems. The area of a rectangle is given by the product of its length and width, represented by the formula \( A = l \times w \). Area measurement is all about covering a surface, quantified in square units such as square centimeters in this case.On the other hand, the perimeter refers to the total distance around the edge of the rectangle. The formula to find the perimeter \( P \) is \( 2(l + w) \), a result of adding the lengths of all four sides together. Rearranging this formula can easily yield a linear equation that relates the length and width.

• Area formula: \( A = l \times w \)
• Perimeter formula: \( P = 2(l + w) \)

These formulas work hand in hand. Given the area and perimeter, we can deduce relationships like \( l + w = 16 \) and find the exact dimensions required by substituting these relationships back into the equations we form from these core formulas. This understanding forms the backbone of solving not only rectangle problems but various geometric puzzles.

Solving Equations

Equation solving is a fundamental skill, central to untangling the values of unknowns in math problems. In our rectangle exercise, solving equations involves utilizing both linear equations derived from perimeter and quadratic equations from the area.One starts by expressing one variable in terms of the other, such as \( l = 16 - w \) from the simplified perimeter equation. This substitution strategy transforms the problem into an equation purely in terms of one variable. Substitution into the area equation \( (16 - w)w = 63 \) leads us to a quadratic equation \( w^2 - 16w + 63 = 0 \).

• Isolate one variable: Begin with simpler (linear) equations.
• Substitute: Replace one variable with its corresponding expression.
• Solve: Use algebraic methods or formulas, like the quadratic formula, to find all possible values.

The ultimate goal is to find numerical values for \( l \) and \( w \) that satisfy both the area and perimeter equations. Solving equations is not merely number crunching—it's about logical manipulation of mathematical relationships to find solutions within given constraints.