From each corner of a square with an edge of 4 cm, a 1 cm segment was chopped off from each side containing a vertex. What is the perimeter and the area of the octagon thus created?
Finding the Perimeter and Area of the Octagon
This problem asks us to find the perimeter and area of a shape created by cutting off the corners of a square. We start with a square that has an edge length of 4 cm. From each of its four corners, a 1 cm segment is 'chopped off' from each side containing a vertex. This process transforms the square into an octagon.
Understanding the Geometry of the Octagon
Let's visualize what happens at each corner of the square. The square has 4 vertices. At each vertex, we take the two sides that meet there. A 1 cm segment is measured along each of these sides starting from the vertex. A cut is made connecting the endpoints of these two 1 cm segments. Since the original square corner is a 90-degree angle, the cut creates a right-angled triangle at each corner with legs of length 1 cm.
When these four corner triangles are removed, the original four sides of the square are shortened, and four new sides are created by the cuts. The resulting shape has 4 sides from the original square's edges and 4 new sides from the cuts, totaling 8 sides, which makes it an octagon.
Calculating the Perimeter of the Octagon
The octagon has 8 sides. Let's determine the length of each side:
- Original sides: The original square sides were 4 cm long. At each end of an original side, a 1 cm segment is removed. So, the length of the remaining part of each original side is the total length minus the two removed segments: \(4 \text{ cm} - 1 \text{ cm} - 1 \text{ cm} = 2 \text{ cm}\). There are four such sides in the octagon.
- New sides (from cuts): These are the hypotenuses of the four right-angled triangles that were removed. Each triangle has legs of length 1 cm. Using the Pythagorean theorem (\(a^2 + b^2 = c^2\)), the length of the hypotenuse (the new side) is \(\sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \text{ cm}\). There are four such sides in the octagon.
The perimeter of the octagon is the sum of the lengths of all its sides:
Perimeter = (Sum of lengths of the four remaining original sides) + (Sum of lengths of the four new sides)
Perimeter = \(4 \times (2 \text{ cm}) + 4 \times (\sqrt{2} \text{ cm})\)
Perimeter = \(8 \text{ cm} + 4\sqrt{2} \text{ cm}\)
Perimeter = \((8 + 4\sqrt{2}) \text{ cm}\)
Calculating the Area of the Octagon
The area of the octagon is the area of the original square minus the total area of the four triangles removed from the corners.
- Area of the original square: The square has an edge length of 4 cm. Area of square = \(\text{side} \times \text{side} = 4 \text{ cm} \times 4 \text{ cm} = 16 \text{ cm}^2\).
- Area of each removed triangle: Each removed triangle is a right-angled triangle with legs of length 1 cm. Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \text{ cm} \times 1 \text{ cm} = 0.5 \text{ cm}^2\).
- Total area of the four removed triangles: Total removed area = \(4 \times (0.5 \text{ cm}^2) = 2 \text{ cm}^2\).
The area of the octagon is:
Area of octagon = Area of square - Total area of removed triangles
Area of octagon = \(16 \text{ cm}^2 - 2 \text{ cm}^2 = 14 \text{ cm}^2\)
Summary of Calculations
| Property | Calculation | Result |
|---|---|---|
| Side length of original square | 4 cm | |
| Length of removed segment at each corner | 1 cm | |
| Length of remaining part of original side | \(4 - 1 - 1\) | 2 cm |
| Number of remaining original sides | 4 | |
| Length of new side (hypotenuse) | \(\sqrt{1^2 + 1^2}\) | \(\sqrt{2}\) cm |
| Number of new sides | 4 | |
| Perimeter of octagon | \(4 \times 2 + 4 \times \sqrt{2}\) | \((8 + 4\sqrt{2})\) cm |
| Area of original square | \(4 \times 4\) | 16 cm\(^2\) |
| Area of one removed triangle | \(\frac{1}{2} \times 1 \times 1\) | 0.5 cm\(^2\) |
| Total area of 4 removed triangles | \(4 \times 0.5\) | 2 cm\(^2\) |
| Area of octagon | \(16 - 2\) | 14 cm\(^2\) |
The calculated perimeter is \((8 + 4\sqrt{2}) \text{ cm}\) and the area is \(14 \text{ cm}^2\).
Revision Table: Square to Octagon Transformation
| Shape | Original Square | Resulting Octagon |
|---|---|---|
| Side Length | 4 cm (edge) | 4 sides of 2 cm, 4 sides of \(\sqrt{2}\) cm |
| Number of Vertices | 4 | 8 |
| Transformation | 4 corners removed (1x1 right triangles) | |
| Perimeter Calculation | \(4 \times 4 = 16\) cm | \(4 \times 2 + 4 \times \sqrt{2} = (8 + 4\sqrt{2})\) cm |
| Area Calculation | \(4 \times 4 = 16\) cm\(^2\) | \(16 - 4 \times (\frac{1}{2} \times 1 \times 1) = 16 - 2 = 14\) cm\(^2\) |
Additional Information: Properties of Octagons
An octagon is an eight-sided polygon. In this specific problem, the octagon created is a special type. Since it is derived from a square by cutting off identical right-angled triangles from the corners, the remaining four longer sides are equal in length (2 cm), and the four shorter sides created by the cuts are also equal in length (\(\sqrt{2}\) cm). The angles of this octagon alternate between the original square's angles (which are now interior to the remaining straight segments, 135 degrees) and the angles at the vertices of the new sides (which are formed by two hypotenuses meeting, 90 degrees). This type of octagon, with alternating equal sides and alternating equal angles, is sometimes referred to as a truncated square.
Understanding how removing parts of a shape affects its perimeter and area is a key concept in geometry. Perimeter is the distance around the edge, so cutting off corners changes the boundary. Area is the space enclosed, so removing parts reduces the area.
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