Two adjacent sides of a rhombus form a $60$-degree angle. If each side of the rhombus measures $2$ cm, what is the area of the rhombus, in square centimeters? Express your answer in simplest radical form.
In rhombus $ABCD$, the acute angle $DAB$ has a measure of $60^\circ$. We drop a perpendicular from $D$ to $\overline{AB}$, which creates a 30-60-90 right triangle. Since the hypotenuse $\overline{AD}$ has a length of $2$ cm, the length of $\overline{AE}$ is $\frac{AD}{2}=1$ cm and the length of $\overline{DE}$ is $AE\cdot\sqrt{3}=\sqrt{3}$ cm. We now know the base of the rhombus is $2$ cm and the height of the rhombus is $\sqrt{3}$ cm, so the area is $bh=\boxed{2\sqrt{3}}$ sq cm.

[asy]/* size(50); import three; defaultpen(linewidth(0.7)); currentprojection = orthographic(1,-2,1/2);

*/ size(100); defaultpen(linewidth(0.7)); real sx = 0.6, sy = 0.2;
path f1 = (0,0)--(1,1.7)--(3,1.7)--(2,0)--cycle;
filldraw(f1, rgb(0.9,0.9,0.9));
path f2=(1,1.7)--(1,0);
draw(f2);
label("$A$",(0,0),SW);
label("$B$",(2,0),SE);
label("$C$",(3,1.7),NE);
label("$D$",(1,1.7),NW);
label("$E$",(1,0),S);

[/asy]