For the three identical squares shown, points $A$, $B$ and $C$ are vertices, and $AB$ = $2\sqrt{5}$ cm. What is the length of $AC$, in centimeters? Express your answer in simplest radical form.

[asy]
draw((0,0)--(0,10)--(30,10)--(30,0)--cycle);
draw((10,0)--(10,10));
draw((20,0)--(20,10));
draw((0,10)--(20,0),linewidth(1));
label("$A$",(0,10),NW);
label("$B$",(20,0),S);
label("$C$",(30,0),S);
[/asy]
Let the side length of one of the squares be $x$. Looking at the right triangle with hypotenuse $AB$, we have the equation $x^2+(2x)^2=(2\sqrt{5})^2$ from the Pythagorean Theorem.  Simplifying this equation gives $x^2=4$. Looking at the right triangle with hypotenuse $AC$, we have the equation $x^2+(3x)^2=AC^2 \Rightarrow AC^2=10x^2=40$. Thus, $AC=\sqrt{40}=\boxed{2\sqrt{10}}$ centimeters.