Area of square formed inside regular octagon

In a regular octagon with side length $s = 6$ cm, each interior angle is $\frac{(8-2)\times 180^\circ}{8} = 135^\circ$.

In $\triangle ABC$, $AB = BC = 6$ and $\angle ABC = 135^\circ$. By Law of Cosines on $\triangle ABC$: $AC^2 = 6^2 + 6^2 - 2(6)(6)\cos(135^\circ)$ $AC^2 = 72 - 72\left(-\frac{1}{\sqrt{2}}\right) = 72 + 36\sqrt{2} = 36(2 + \sqrt{2})$

Since $ACEG$ is a square formed by alternate vertices, its area is $AC^2 = 36(2 + \sqrt{2})$ sq cm.