Problem 45 - Precalculus: Mathematics for Calculus

Question

A rectangle has an area of \(180 \mathrm{cm}^{2}\) and a perimeter of \(54 \mathrm{cm} .\) What are its dimensions?

Step-by-Step Solution

Answer

The dimensions of the rectangle are 15 cm by 12 cm.

Step 1: Define Variables

Let's denote the length of the rectangle as \( l \) and the width as \( w \). The formulas for the area \( A \) and perimeter \( P \) are given by \( A = l \times w \) and \( P = 2(l + w) \).

Step 2: Write Equations

From the problem, we know the area is 180 \( \text{cm}^2 \) and the perimeter is 54 cm. We can set up the following equations: \( l \times w = 180 \) and \( 2(l + w) = 54 \).

Step 3: Simplify Perimeter Equation

Simplify the perimeter equation: \( l + w = 27 \). Divide both sides of the perimeter equation \( 2(l+w)=54 \) by 2.

Step 4: Substitute and Solve

From the simplified perimeter equation \( l + w = 27 \), solve for one variable in terms of the other, such as \( l = 27 - w \). Substitute this into the area equation: \((27 - w) \times w = 180\).

Step 5: Solve Quadratic Equation

Expand and set the area equation in quadratic form: \( 27w - w^2 = 180 \) becomes \( w^2 - 27w + 180 = 0 \). This is a quadratic equation.

Step 6: Factorize Quadratic Equation

Factorize the quadratic equation \( w^2 - 27w + 180 = 0 \) to \( (w - 15)(w - 12) = 0 \).

Step 7: Solve for Width

Set each factor to zero and solve for \( w \): \( w = 15 \) or \( w = 12 \).

Step 8: Determine Length

Use \( l = 27 - w \). If \( w = 15 \), then \( l = 12 \). If \( w = 12 \), then \( l = 15 \). Therefore, the dimensions are 15 cm and 12 cm.

Key Concepts

Understanding Area

In geometry, understanding the concept of "area" is crucial when dealing with shapes. The area is defined as the measure of the surface contained within a shape. For rectangles, a common geometric figure, the formula used is straightforward:

• Area of a rectangle = Length × Width (\( A = l \times w \)).

This means you multiply the length of one side by the length of the adjacent side.
Consider measuring the carpet needed to cover a room - the carpet required would be determined by the area of that room. If you know two of the parameters (in this case, the area and one dimension), you can solve for the unknown dimension. In problems like our exercise, knowing the area allows us to set up equations that aid in finding the shape's dimensions.

Calculating Perimeter

The perimeter of a shape refers to the total distance around the shape. It's like the path you would trace if you walked around the entire boundary. For a rectangle, the perimeter can be calculated using this formula:

• Perimeter of a rectangle = 2 × (Length + Width) (\( P = 2(l + w) \)).

This formula comes from the fact that a rectangle has two lengths and two widths. Perimeter problems often ask for missing dimensions or the space around a shape, making it an essential concept in geometry.
In our exercise, understanding perimeter helped set up the equation to simplify the task of finding the dimensions. When combined with the area, these fresh perspectives provide two simultaneous equations you'd need to solve, leading us to discover the actual size and dimension of the geometric shape.

Solving Quadratic Equations

Quadratic equations are a type of polynomial equation that includes a variable raised to the second power. They take the form:

• \( ax^2 + bx + c = 0 \)

where \( a, b, \) and \( c \) are constants, and \( x \) is the variable.
Solving these can involve various methods like factoring, using the quadratic formula, or completing the square. In the solved exercise, factoring was used where the quadratic equation was expressed as:

• \( w^2 - 27w + 180 = 0 \)

This equation was factored into \((w - 15)(w - 12) = 0 \).
By setting each factor to zero, you find the solutions \( w = 15 \) or \( w = 12 \). Quadratic equations are foundational in algebra and appear widely in various mathematical applications, making them vital for solving problems related to areas and perimeters in geometry.