Perimeter of rhombus

Let $x$ be the side length of the rhombus. The area is given by $x^2 \sin \theta$ where $\theta$ is the included angle. So we have

$800\sqrt{3}=x^2 \sin~60$

$800\sqrt{3}=x^2\left(\dfrac{\sqrt{3}}{2}\right)$

$1600=x^2$

$x=40$

Thus, the desired perimeter is $4\times 40=\boxed{160~cm}$

We look at $\displaystyle AB$ and $\displaystyle BC$ not as mere sides of a rhombus but as vectors.

$Area(\Box ABCD) = | \vec{BA} \times \vec{BC} |$

$\displaystyle \Rightarrow 800 \sqrt3 = |\vec{BA} |\cdot |\vec{BC} |\cdot \sin( 60^{\circ})$

But $\displaystyle |\vec{BA} | = |\vec{BC} |$ since, sides of a rhombus are equal. Substituting this, we get:

$\displaystyle \Rightarrow |\vec{BA} | = 40$

$\displaystyle Perimeter = 4 \times |\vec{BA} | = \boxed{160 }$