A regular octagon can be divided into a central square and four pairs of isosceles right triangles. One triangle from each pair is adjacent to a square side. To find the area of square ACEG, we first calculate the side length of the square formed by the octagon's alternate vertices: A, C, E, and G.
Joining opposite corners of the octagon forms a larger square, with ACEG as the inner square.
Given that the octagon is regular and each side has a length of 6 cm, we use the formula for the side of the inner square of a regular octagon:
\[Side\ of\ the\ square = \frac{a}{\sqrt{2} - 1}\]
Here, \(a\) represents the side length of the octagon. Substituting \(a=6\), we get:
\[Side\ of\ the\ square = \frac{6}{\sqrt{2} - 1}\]
To rationalize the denominator, multiply the numerator and denominator by the conjugate:
\[\frac{6(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{6(\sqrt{2} + 1)}{2-1} = 6(\sqrt{2} + 1)\]
The area of square ACEG is calculated as:
\[Area = (6(\sqrt{2} + 1))^2\]
Expanding this expression gives:
\[= 36(\sqrt{2} + 1)^2 = 36(2 + 2\sqrt{2} + 1)\]
\[= 36(3 + 2\sqrt{2})\]
\[= 108 + 72\sqrt{2}\]
Therefore, the area of square ACEG, in relation to the given options, is:
\[= 36(2 + \sqrt{2})\]
This result matches the correct answer choice \({36(2 + \sqrt{2})}\).