We can solve this problem by using the Pythagorean theorem and the properties of a regular hexagon.

First, let's draw a diagram:

[asy]
pair A,B,C,D,E,F,G;
A = (0,0);
B = (1,0);
C = (1.5,sqrt(3)/2);
D = (1,2*sqrt(3)/2);
E = (0,2*sqrt(3)/2);
F = (-0.5,sqrt(3)/2);
G = (0.5,sqrt(3)/2);
draw(A--B--C--D--E--F--cycle);
draw(A--C);
draw(G--D,dashed);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NE);
label("$E$",E,NW);
label("$F$",F,W);
label("$G$",G,NE);
[/asy]

We know that $AC$ is a side of the square, so its length is also 1 cm. Let's find the length of $GC$ using the Pythagorean theorem. Since $\triangle GAC$ is a 30-60-90 triangle, we have $GC = AC\sqrt{3} = \sqrt{3}$.

Next, we can use the fact that $ABCDEF$ is a regular hexagon to find the area. We can split the hexagon into 6 equilateral triangles with side length 1 cm:

[asy]
pair A,B,C,D,E,F,G;
A = (0,0);
B = (1,0);
C = (1.5,sqrt(3)/2);
D = (1,2*sqrt(3)/2);
E = (0,2*sqrt(3)/2);
F = (-0.5,sqrt(3)/2);
G = (0.5,sqrt(3)/2);
draw(A--B--C--D--E--F--cycle);
draw(A--C);
draw(G--D,dashed);
draw(A--C--D--cycle,red);
draw(B--F--E--cycle,red);
draw(C--G--F--cycle,red);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NE);
label("$E$",E,NW);
label("$F$",F,W);
label("$G$",G,NE);
[/asy]

Each of these triangles has area $\frac{1}{2}\cdot 1\cdot 1\cdot \sqrt{3} = \frac{\sqrt{3}}{2}$. Therefore, the area of the hexagon is $6\cdot \frac{\sqrt{3}}{2} = 3\sqrt{3}$.

Finally, we can find the area of the square with side $AC$ using $A = s^2$. Therefore, the area of the square is $1\cdot 1 = 1$ square cm.

To summarize:

- The area of the regular hexagon is $3\sqrt{3}$ square cm.
- The area of the square with side $AC$ is 1 square cm.

Therefore, the answer is $\boxed{1}$.