Basic Geometry

The area cut off on each corner has to be a right angled isosceles triangle to retain the property of a regular polygon. Let the length cut off from any one side at the corner be $x$ . Then, using Pythagorean theorem, the side of the octagon would be $\sqrt{x^2+x^2}$ . Similarly, the side of the octagon would also be $2-2x$ (Since we cut off $2x$ length from each side. $\sqrt{2} x = 2 - 2x \Rightarrow x= 2-\sqrt{2}$ Area of each triangle cut off will be $\frac 12 \times x \times x$ and 4 such triangles would be cut off, so the remaining area would be : $4 - \frac{4(2- \sqrt{2})^2}{2} \approx 3.31$

Let the length of side of the octagon be $a$ cm, then we know that:

$a + 2\left( \dfrac {a}{\sqrt{2}} \right) = 2 \quad \Rightarrow a + \sqrt{2}a = 2\quad \Rightarrow a = \dfrac {2} {1+\sqrt{2}} = 2(\sqrt{2}-1)$

We also know that the area cut out from 2cm-by-2cm square is:

$4 \left( \frac {1}{2} \right) \left( \frac {a}{\sqrt{2}} \right) \left( \frac {a}{\sqrt{2}} \right) = a ^2$

Therefore the area of the octagon

$A = 4 - a^2 = 4 - 2^2(\sqrt{2}-1)^2 = 4 - 4(2 - 2\sqrt{2} + 1)$ $\quad = - 8 + 8\sqrt{2} = 8(\sqrt{2}-1) = \boxed {3.31}$

The formula for the area of a regular polygon is $A=nr^{2} \tan (\frac{\pi}{n})$ In this formula, $n$ is the number of sides and $r$ is the apothem; that is, the distance from the midpoint of a side to the center. Given we have an octagon converted from a square with side length 2 cm, it is implied that the apothem of the octagon is 1 cm. So plugging in: $A=(8)(1)^{2} \tan (\frac{\pi}{8})=8 \tan (\frac{\pi}{8})$ But we know that $\frac{\pi}{8}$ is half of $\frac{\pi}{4}$ , so we can rewrite: $A=8 \frac{\sec (\frac{\pi}{4})-1}{\tan (\frac{\pi}{4})}=8(\frac{\sqrt{2}-1}{1}) \approx \boxed{3.31}$

Let octagon side be a and left over x now 2x²=a²,a=xroot2 a+2x=2 from this we get x=2-root2 area of octagon= area of square-area of triangles thus formed. Area of octagon=4-2x²=8(root2-1)=3.31

Let side of octagon be 'x'......so using pythagoras...sid of triangle at corner is x/(root2)....so using simple geometry ....xroot2+x=2.............so,x=2 (root2 -1)...............area of triangle of 1 corner is half×base ×height...i.e. half×(x/root2)^2.....total area of triangles at corners will become 4times above area...=4 (root2-1)^2.............we have to find area of octagon so subtract this area from area of square i.e.4 cm^2...... (Put root2=1.414) So ans is 3.312 but ,we want upto 2 decimal so ans is 3.31