PQRS is a square whose side length is 16 cm. A largest possible regular octagon is formed by cutting equal isosceles right triangles from each corner of the square. What is the length of the side of this regular octagon in centimetres?

Introduction / Context:
This problem asks for the side length of the largest regular octagon that can be cut from a square by removing equal isosceles right triangles at each corner. This is a classic geometry configuration. The side of the octagon is related in a fixed way to the side of the square, and the relationship involves the square root of 2 due to the geometry of 45 degree angles and diagonals in the square.

Given Data / Assumptions:

• PQRS is a square with side length 16 cm.
• Equal isosceles right triangles are cut off at each of the four corners.
• The remaining central shape is a regular octagon.
• The octagon is the largest that can be inscribed in this way.
• We need the side length of this regular octagon.

Concept / Approach:
Let the side of the square be a and let the length cut off along each side be x. Then each corner triangle has legs of length x, and the cut removes x from each adjacent side. The middle portion between two consecutive cuts forms one side of the octagon. The remaining length on each original side is a − 2x. The sloping edges are formed along the diagonals of the corner triangles, which are of length x√2, but that diagonal becomes part of the octagon side in a symmetrical configuration. For the largest regular octagon, the octagon side s is equal to the remaining straight part between the cuts, and there is a known relation s = a(√2 − 1).

Step-by-Step Solution:
Step 1: Let the side of the square be a. Here a = 16 cm.Step 2: Consider one corner where a right isosceles triangle is cut. If the perpendicular legs along the square sides are x and x, then the hypotenuse, which becomes one side of the octagon, has length x√2.Step 3: After cutting, the straight portion of the original side between adjacent cut points has length a − 2x.Step 4: For a regular octagon inscribed in this way, the side length s of the octagon satisfies s = a(√2 − 1). This standard result can be derived by equating the geometry of the diagonals and the equal side lengths but is often memorised for such questions.Step 5: Substitute a = 16 to obtain s = 16(√2 − 1) = 16√2 − 16 cm.

Verification / Alternative check:
To double check, note that if a = 1, the formula gives s = √2 − 1, which matches the derived relation in many geometry texts for a unit square. Here, with a scale factor 16, all linear dimensions scale by 16, so the octagon side becomes 16(√2 − 1). Numerically, √2 is about 1.414, so s ≈ 16(1.414 − 1) = 16 * 0.414 ≈ 6.624 cm, which is a reasonable side length less than 16 cm, consistent with the octagon fitting inside the square.

Why Other Options Are Wrong:

• 8√2 − 4 cm and 8(√2 − 1) cm correspond to a square of side 8 cm, not 16 cm.
• 4(2√2 − 1) cm is another incorrect scaling that does not fit the standard formula for this construction.
• 16 − 8√2 cm is negative when computed numerically, which is impossible for a side length.

Common Pitfalls:

• Assuming that the octagon side equals the remaining straight length a − 2x without relating x to a using right triangle geometry.
• Mixing up the factor √2 − 1 with 2 − √2 or other similar looking expressions.
• Trying to compute everything from scratch without using the known result s = a(√2 − 1), which can lead to algebra errors.

Final Answer:
The side length of the largest regular octagon that can be cut from the square is 16√2 − 16 cm.