Problem 12 - Basic Engineering Mathematics

Question

The area of a rectangle is \(23.6 \mathrm{~cm}^{2}\) and its width is \(3.10 \mathrm{~cm}\) shorter than its length. Determine the dimensions of the rectangle, correct to 3 significant figures.

Step-by-Step Solution

Answer

Length: 6.65 cm, Width: 3.55 cm.

Step 1: Define Variables

Let the length of the rectangle be denoted as \( l \) and the width as \( w \). We know that the width is 3.10 cm shorter than the length. So, we can express the width as \( w = l - 3.10 \).

Step 2: Write the Area Equation

The formula for the area of a rectangle is \( ext{Area} = ext{length} \times ext{width} \). We are given that the area is \( 23.6 \, \text{cm}^2 \), so we set up the equation \( l \times w = 23.6 \). Substituting the expression for \( w \), we have \( l(l - 3.10) = 23.6 \).

Step 3: Simplify the Equation

Expand the equation \( l(l - 3.10) = 23.6 \) to \( l^2 - 3.10l = 23.6 \). This is now a quadratic equation in terms of \( l \).

Step 4: Solve the Quadratic Equation

Re-arrange the quadratic equation to standard form: \( l^2 - 3.10l - 23.6 = 0 \). Use the quadratic formula \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -3.10 \), and \( c = -23.6 \). Calculate the discriminant \( b^2 - 4ac = (-3.10)^2 - 4 \times 1 \times (-23.6) \), which simplifies to \( 9.61 + 94.4 = 104.01 \).

Step 5: Calculate Length Solutions

Substitute the discriminant back into the quadratic formula: \[ l = \frac{-(-3.10) \pm \sqrt{104.01}}{2 \times 1} \]. Calculate the roots: \( l = \frac{3.10 \pm 10.2}{2} \). The relevant positive solution is \( l = \frac{13.3}{2} = 6.65 \) cm.

Step 6: Calculate Width

Using the relation \( w = l - 3.10 \), substitute the value of \( l = 6.65 \). Hence, \( w = 6.65 - 3.10 = 3.55 \) cm.

Key Concepts

Area of a rectangle

Understanding the area of a rectangle is crucial in geometry and mathematics. The area is simply the space within the perimeter of the rectangle. It's calculated by multiplying the length by the width.
In the given exercise, we are provided with an area of 23.6 cm². This value represents the total space within the rectangle.
Knowing one dimension, like the width or length, allows us to rearrange and solve for the other dimension using the area formula \[ \text{Area} = l \times w \].
Real-world application includes measuring land or any rectangular surfaces. Remember, always use consistent units when dealing with area calculations.

Quadratic equation

A quadratic equation is a polynomial equation in the form of \( ax^2 + bx + c = 0 \). It is characterized by the highest degree of variable being two.
In our problem, the equation \( l^2 - 3.10l - 23.6 = 0 \) was derived from the rectangle area setup. This implies the length \( l \) satisfies a quadratic expression.
Quadratic equations arise in various mathematical contexts, and solving them typically involves methods like factoring, completing the square, or applying the quadratic formula.

• Factoring: Rearrange and express the equation as a product of two binomials.
• Completing the Square: Rewrite the equation in a perfect square form.
• Quadratic Formula: \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), a universal method useful when other methods are infeasible.

Understanding when and how to apply these solutions is an essential skill.

Solving equations

Solving equations is about finding the values of the variables that satisfy the equation. This exercise focuses on solving a quadratic equation for the rectangle dimensions.
The main approach used here is the quadratic formula method. Its primary advantage is being universally applicable to all quadratic equations, ensuring we find solutions even when others, like factoring, might not work.
After establishing the quadratic form, evaluating the discriminant \(b^2 - 4ac\) is crucial. It reveals whether the solutions are real or complex.

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is exactly one real solution.
• If \( b^2 - 4ac < 0 \), the solutions are complex and not applicable for real-world length calculation.

Applying this method, our exercise solved for a real and applicable length of the rectangle.

Geometry in mathematics

Geometry is a branch of mathematics concerned with the properties and relations of points, lines, surfaces, and solids. In this exercise, the focus is on the properties of rectangles.
A rectangle is a four-sided figure with opposite sides equal and all angles right-angled (90 degrees). Understanding these properties allows mathematical modeling of real-world structures.
The principles of geometry help us derive equations based on spatial configurations and inform us how to approach solving them using algebraic methods.

• Recognize shapes and their properties to set up accurate equations.
• Use spatial reasoning to visualize problems.
• Apply equations to determine dimensions and other characteristics effectively.

Geometry's intertwining with algebra helps not just in theoretical problems but also in engineering, architecture, and various fields where spatial measurements are vital.