Problem 41 - Algebra and Trigonometry

Question

Dimensions of a Rectangle 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 rectangle's dimensions are 15 cm by 12 cm.

Step 1: Define Variables

Let the length of the rectangle be \( l \) and the width be \( w \). We have two equations to work with: the area equation and the perimeter equation.

Step 2: Area Equation

The area \( A \) of a rectangle is given by \( A = l \times w \). According to the problem, the area is 180 \( \mathrm{cm}^2 \). Therefore, we have the equation: \[ l \times w = 180 \]

Step 3: Perimeter Equation

The perimeter \( P \) of a rectangle is given by \( P = 2l + 2w \). According to the problem, the perimeter is 54 \( \mathrm{cm} \). Therefore, we have the equation: \[ 2l + 2w = 54 \] This can be simplified to: \[ l + w = 27 \]

Step 4: Solve System of Equations

We now have the following two equations:1. \( l \times w = 180 \)2. \( l + w = 27 \)We can express \( l \) in terms of \( w \) from the second equation:\[ l = 27 - w \] Substituting in the first equation, we get:\[ (27 - w) \times w = 180 \]\[ 27w - w^2 = 180 \]\[ w^2 - 27w + 180 = 0 \]

Step 5: Solve Quadratic Equation

Now solve the quadratic equation \( w^2 - 27w + 180 = 0 \). To do this, we can use the quadratic formula where \( a = 1 \), \( b = -27 \), and \( c = 180 \): \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute the values:\[ w = \frac{-(-27) \pm \sqrt{(-27)^2 - 4 \times 1 \times 180}}{2 \times 1} \]\[ w = \frac{27 \pm \sqrt{729 - 720}}{2} \]\[ w = \frac{27 \pm 3}{2} \]So, \( w = 15 \) or \( w = 12 \).

Step 6: Determine Corresponding Lengths

For \( w = 15 \), use \( l = 27 - w \) to find \( l \):\[ l = 27 - 15 = 12 \]For \( w = 12 \), again use \( l = 27 - w \):\[ l = 27 - 12 = 15 \]Thus, the dimensions are \( 15 \text{ cm} \times 12 \text{ cm} \).

Key Concepts

Understanding Quadratic Equations

A quadratic equation is a mathematical expression of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. The highest exponent in this kind of equation is two, hence the name 'quadratic'. Solving quadratic equations is fundamental in algebra, often involving finding the values of \( x \) (known as roots) that make the equation true.
The most common method to find these solutions is the quadratic formula:

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

This formula involves calculating the discriminant \( b^2 - 4ac \), which helps in determining the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root.
• If negative, the roots are complex.

In our rectangle problem, the quadratic equation \( w^2 - 27w + 180 = 0 \) arises from expressing the rectangle's dimensions in terms of one variable. Solving this provides possible widths for the rectangle.

Calculating the Area of a Rectangle

The area of a rectangle is a useful calculation in geometry that helps us understand the size of a two-dimensional space. To find the area, you multiply the length \( l \) by the width \( w \). This can be expressed with the formula:

• \( A = l \times w \)

The units of area are always squared, representing the number of square units that fit inside the rectangle.
For example, when given that the area is 180 \( \text{cm}^2 \) as in our rectangle problem, it means if you slice the rectangle into small, 1cm by 1cm squares, you'd perfectly fit 180 of them.

• This property is crucial when determining one dimension if you know the area and another dimension.

In exercises where both area and perimeter data are provided, you can create simultaneous equations to find both unknowns, which are particularly useful for calculating dimensions like length and width.

Determining the Perimeter of a Rectangle

The perimeter of a rectangle is a linear measurement that tells us the distance around the outside of the rectangle. It is measured in linear units such as centimeters or meters, not squared like the area.
To calculate the perimeter, you add two times the length to two times the width. The formula is:

• \( P = 2l + 2w \)

In practical terms, the perimeter reflects the amount of material needed to enclose the rectangle fully, like a fence around a yard. For the problem given, the perimeter is 54 cm, meaning it takes 54 cm of material to go around the rectangle.
If you know one dimension and the perimeter, you can rearrange the formula to solve for the other dimension:

• For instance, \( l + w = 27 \) is derived from \( 2l + 2w = 54 \).

Understanding this concept helps in identifying and calculating the unknown lengths or widths of rectangles in geometric problems.