Mathematical Geometry: Exploring the Perimeter and Area of an Octagon Formed from a Square

The problem of transforming a square into an octagon by cutting off segments from each corner is a fascinating exercise in geometry. In this article, we will provide a detailed analysis, step-by-step solution, and practical insights into how to determine the perimeter and area of the resulting octagon.

Step 1: Analyzing the Original Square

Consider a square with an edge length of 4 cm. This square will be our starting point.

Perimeter: The perimeter of a square is given by the formula: t[text{Perimeter} 4 times text{side length} 4 times 4 16 text{ cm}] Area: The area of a square can be calculated using the formula: t[text{Area} text{side length}^2 4^2 16 text{ cm}^2]

Step 2: Understanding the Modification

When we remove a 1 cm segment from each corner of the square, we are essentially cutting off small right triangles from each corner. Each of these triangles has legs of 1 cm in length, resulting in an octagon.

Area of One Triangle: The area of one of these right triangles can be calculated as follows: t[text{Area of one triangle} frac{1}{2} times text{base} times text{height} frac{1}{2} times 1 times 1 0.5 text{ cm}^2] Total Area Removed: Since there are four corners, the total area removed from the square is: t[text{Total area removed} 4 times 0.5 2 text{ cm}^2]

Step 3: Calculating the Area of the Octagon

The area of the octagon is simply the area of the original square minus the area of the triangles that were removed:

t[text{Area of the octagon} text{Area of the square} - text{Total area removed} 16 - 2 14 text{ cm}^2]

Step 4: Calculating the Perimeter of the Octagon

Each side of the original square loses 1 cm from each end, resulting in each of the straight sides of the octagon being:

t[text{New side length} 4 - 2 times 1 2 text{ cm}]

The octagon has 4 sides of 2 cm each, plus the hypotenuses of the right triangles that were removed from the corners. Each hypotenuse is the same as the side of the removed triangles, which are 1 cm each.

Thus, the perimeter of the octagon is:

t[text{Perimeter} 4 times 2 4 times 1 8 4 12 text{ cm}]

Final Results

Perimeter of the Octagon: 12 cm Area of the Octagon: 14 cm2

This exercise not only highlights the beauty of geometric transformations but also demonstrates the practical application of mathematical principles. Understanding how to manipulate shapes and calculate their properties is essential in various fields, from art and design to engineering and architecture.