Properties of Hexagon and Octagon

Area of a regular hexagon = $$\frac{3\sqrt{3}}{2}x^2$$

Area of an equilateral triangle = $$\frac{\sqrt{3}}{4}\left(a\right)^2$$ ; where a = side of the triangle

Since the area of the two figures are equal, we can equate them as folllows: $$\frac{3\sqrt{3}}{2}x^2=\frac{\sqrt{3}}{4}\left(12\right)^2$$

On simplifying: $$x^2=24\ $$

$$\therefore\ x=2\sqrt{6}$$

Since a regular hexagon can be considered to be made up of 6 equilateral triangles, a line joining the farthest vertices of a hexagon can be considered to be made up using the sides of two opposite equilateral triangle forming the hexagon. Hence, its length should be twice the side of the hexagon, in this case, 4 cm.

Now, AD divided the hexagon into two symmetrical halves. Hence, AD bisects angle D, and hence, angle ADC is $$60^{\circ\ }$$.

We can find out the value of AT using cosine formula:

$$AT^2=\ 4^2+1^2-2\times\ 1\times\ 4\cos60$$

$$AT^2=\ 17-4=13$$

$$AT=\sqrt{\ 13}$$

This is the figure in the question, 

We are given that each side is 6cm long, 
To find the side AC, we can use cosine rule, since we know each interior angle of the octagon(which is 135 degrees)

$$\cos\left(\angle ABC\right)=\dfrac{\left(AB^2+BC^2-AC^2\right)}{2\left(AB\right)\left(AC\right)}$$

$$\cos\left(135\right)=\dfrac{\left(36+36-AC^2\right)}{2\left(36\right)}$$

$$-\dfrac{1}{\sqrt{2}}=\dfrac{\left(72-AC^2\right)}{72}$$

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

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

Since AC is the side of the square, and the area of a square is square of the side. 

Answer is $$36\left(2+\sqrt{2}\right)$$