A regular octagon ABCDEFGH has sides of length 6 cm each. Then, the area, in square cm, of the square ACEG is:

The correct option is 36(2 + $\sqrt{2}$).

Step 1: Understand the properties of a regular octagon
A regular octagon has eight equal sides of length $s=6\text{cm}$.
The interior angle $\theta$ of a regular polygon with $n$ sides is calculated using the formula:

$\theta =\frac{(n-2)\times 180°}{n}$

For a regular octagon where $n=8$:

$\theta =\frac{(8-2)\times 180°}{8}=\frac{6\times 180°}{8}=135°$

Thus, each interior angle of the octagon, such as angle $\angle \mathrm{ABC}$, is equal to $135°$.

Step 2: Identify the side of the square ACEG
The square $\mathrm{ACEG}$ is formed by connecting alternating vertices $A$, $C$, $E$, and $G$ of the octagon.
The side length of this square is the distance $\mathrm{AC}$.

Step 3: Calculate the length of the diagonal segment $\mathrm{AC}$
Consider the triangle $\mathrm{ABC}$ formed by vertices $A$, $B$, and $C$:
• The side lengths are $\mathrm{AB}=\mathrm{BC}=6\text{cm}$ (sides of the regular octagon).
• The included angle is $\angle \mathrm{ABC}=135°$.
Applying the Law of Cosines to triangle $\mathrm{ABC}$:

${\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2-2\times \mathrm{AB}\times \mathrm{BC}\times \cos (135°)$

Using the trigonometric value $\cos (135°)=-\frac{\sqrt{2}}{2}$, we substitute the known values:

${\mathrm{AC}}^2=6^2+6^2-2\times 6\times 6\times (-\frac{\sqrt{2}}{2})$

Simplify the terms:

${\mathrm{AC}}^2=36+36+36\sqrt{2}$

${\mathrm{AC}}^2=72+36\sqrt{2}$

Factoring out $36$ gives:

${\mathrm{AC}}^2=36(2+\sqrt{2})$

Step 4: Determine the area of square ACEG
The area of the square $\mathrm{ACEG}$ is equal to the square of its side length, which is ${\mathrm{AC}}^2$:

$\text{Area}={\mathrm{AC}}^2=36(2+\sqrt{2}){\text{cm}}^2$