An octagon — eight sides, but not a regular one.
Take the square with side 3, so the marks are 1 unit apart. Each corner cut removes a right isosceles triangle with legs 1, so its hypotenuse is √2.
The 8 sides go: 1, √2, 1, √2, 1, √2, 1, √2
1 ≠ √2 ≈ 1.414, so the sides are not all equal
All 8 angles are equal (135° each)
Area left = 9 − 4 × ½ × 1 × 1 = 9 − 2 = 7 sq. units, i.e. 7⁄9 of the square
Why the triangle gives a regular figure and the square does not: in the triangle the corner angle is 60°, so the corner piece cut off is equilateral and its third side is also 1. In the square the corner angle is 90°, so the piece cut off is right-angled and its hypotenuse is √2 — longer than the sides it replaced. To make the octagon regular you would have to cut at a different distance, not at the thirds.
Try This: Where should you mark a square of side 3 so that the octagon comes out regular? You need the cut length x to satisfy x√2 = 3 − 2x, which gives x = 3⁄(2 + √2) ≈ 0.879 — not one-third.