Problem 79 - Elementary Algebra Within Reach

Question

Geometry The perimeter of a rectangle is 260 meters. The length is 30 meters greater than the width. Find the dimensions of the rectangle.

Step-by-Step Solution

Answer

The dimensions of the rectangle are: Length = 80 meters, Width = 50 meters.

Step 1: Formulate the equations

The perimeter \(P\) of a rectangle can be calculated by the formula \(P = 2l + 2w\), where \(l\) is the length and \(w\) is the width. Given \(P = 260\) m, we get the first equation \[2l + 2w = 260\] Also, we know the length is \(30\) m greater than the width, leading to the second equation \[l = w + 30\]

Step 2: Substitute and Solve

Substitute the second equation into the first to solve for \(w\), \[2(w + 30) + 2w = 260\] which simplifies to \[4w + 60 = 260\] Further simplifying gives \[4w = 200\] Hence, \(w = 50\) m.

Step 3: Determine the length

Substitute \(w = 50\) m into our second equation to determine the length: \[l = 50 + 30 = 80\] m.

Key Concepts

Understanding Rectangle Perimeter

A rectangle is a basic shape in geometry with four sides and opposite sides that are equal in length. To determine its size, the concept of perimeter is useful. The perimeter of a rectangle is the total distance around the rectangle. Simply put, you add up the lengths of all the sides. In mathematical terms, the formula for the perimeter of a rectangle is:

• \[ P = 2l + 2w \]

where \( P \) represents the perimeter, \( l \) represents the length, and \( w \) represents the width. This formula comes from the fact that a rectangle has two lengths and two widths. For the given exercise, with a perimeter of 260 meters, knowing just this one formula allows you to make headway into calculating unknown side dimensions.

Solving Algebraic Equations in Geometry

To find the dimensions of the rectangle, we must solve two algebraic equations simultaneously. Algebraic equation solving involves isolating variables (like \( l \) and \( w \)), to uncover unknown values. In our problem, we start with two primary equations derived from the provided information:

• From the perimeter: \( 2l + 2w = 260 \)
• From the length description: \( l = w + 30 \)

The second equation provides a relationship between length and width, allowing substitution. By inserting \( l = w + 30 \) into the perimeter equation, you can simplify and rearrange terms to solve for the width.Substituting results in a simplified combined equation.

• Substitute: \( 2(w + 30) + 2w = 260 \)
• Simplified: \( 4w + 60 = 260 \)
• Solve to find \( w = 50 \)

Understanding this process is crucial for working out problems where you have multiple conditions to satisfy.

Broader Geometry Concepts and Applications

Geometry is the field of mathematics dealing with shapes, sizes, and dimensions of figures and objects. It includes the study of points, lines, surfaces, solids, and their relationships and properties. In problems like our rectangle scenario, you'll often apply different geometry concepts together. In the case of this exercise:

• You began with a known total perimeter to find individual side dimensions.
• The concept of setting equations based on described relationships allowed you to integrate algebra into geometry.

These principles are incredibly valuable not just for rectangles, but for a wide variety of shapes. By learning and practicing how they interconnect, you gain tools that assist in various real-world applications, such as architecture, engineering, and everyday problem-solving. Understanding the role that algebra and geometry play when they come together makes tackling geometrical problems a feasible and structured task. The ability to dissect a problem, employ formulas, and logically deduce unknowns is at the heart of mathematical studies.