A rectangle has length 8 cm and breadth 6 cm. From each of its four vertices, straight cuts are made so that the remaining figure is a regular octagon (all sides equal). What is the length, in centimetres, of each side of this regular octagon?

Introduction / Context:
In this geometry question, we start with a rectangle whose length is 8 cm and breadth is 6 cm. Each vertex is cut off in such a way that the remaining figure becomes a regular octagon, meaning all eight sides are equal in length. The main challenge is to express the side length of this octagon in terms of the dimensions of the original rectangle and the amounts cut off at the corners.

Given Data / Assumptions:

• Original rectangle: length = 8 cm, breadth = 6 cm.
• Four corners are cut to form a regular octagon.
• Let the amount cut off along the longer side at each corner be x cm.
• Let the amount cut off along the shorter side at each corner be y cm.
• The resulting octagon has all sides equal in length.

Concept / Approach:
The new figure has 8 equal sides. These sides come from:

• The remaining horizontal portions of the length: 8 - 2x.
• The remaining vertical portions of the breadth: 6 - 2y.
• The slant (diagonal) edges formed by cutting off the corners: each diagonal has length √(x^2 + y^2).

For a regular octagon, all these three kinds of edges must have the same length. This gives us equations relating x, y and the octagon side length a, and solving these equations yields the required side length.

Step-by-Step Solution:
Let the side length of the octagon be a. Horizontal side length: a = 8 - 2x. Vertical side length: a = 6 - 2y. Diagonal side length: a = √(x^2 + y^2). From a = 8 - 2x and a = 6 - 2y, we get 8 - 2x = 6 - 2y. So 2 = 2x - 2y → x - y = 1 → x = y + 1. Now a = 6 - 2y. Also a^2 = x^2 + y^2. Substitute x = y + 1: a^2 = (y + 1)^2 + y^2 = 2y^2 + 2y + 1. But a = 6 - 2y, so (6 - 2y)^2 = 2y^2 + 2y + 1. Compute left side: (6 - 2y)^2 = 36 - 24y + 4y^2. Set equal: 36 - 24y + 4y^2 = 2y^2 + 2y + 1. Rearrange: 4y^2 - 2y^2 - 24y - 2y + 36 - 1 = 0 → 2y^2 - 26y + 35 = 0. Divide by 1: 2y^2 - 26y + 35 = 0. Solve quadratic: y = [26 ± √(26^2 - 4 * 2 * 35)] / (2 * 2). Compute discriminant: 26^2 = 676, 4 * 2 * 35 = 280, so √(676 - 280) = √396 = √(36 * 11) = 6√11. Thus y = (26 ± 6√11) / 4 = (13 ± 3√11) / 2. We need y to be less than 3 (so that 6 - 2y > 0), so we choose y = (13 - 3√11) / 2. Now a = 6 - 2y = 6 - (13 - 3√11) = 3√11 - 7.
Verification / Alternative check:
Approximate √11 ≈ 3.316, so 3√11 ≈ 9.948. Then a ≈ 9.948 - 7 = 2.948 cm, which is a reasonable side length (less than both 6 and 8). Also y ≈ (13 - 9.948) / 2 ≈ 1.526 cm, so the trims are small but positive, which is geometrically valid.
Why Other Options Are Wrong:
Option B (5√13 – 8) gives a much larger side length that would not fit inside the original 8 cm by 6 cm rectangle once corners are cut. Option C (4√7 – 11) and Option D (6√11 – 9) either give negative or unrealistic dimensions when checked numerically. Option E (2√10 – 5) does not satisfy the algebraic relationship derived from equating horizontal, vertical and diagonal sides.
Common Pitfalls:
A common mistake is to assume the cuts at each corner are squares (x = y), which only works if the original figure is a square, not a rectangle. Another error is to assume that 8 - 2x automatically equals 6 - 2x, which is impossible and leads to contradiction. Students may also forget to check which root of the quadratic is geometrically valid (positive but small enough so that the remaining sides stay positive).
Final Answer:
Thus, the length of each side of the regular octagon formed from the 8 cm by 6 cm rectangle is 3√11 – 7 centimetres.