Since we have four 30-60-90 triangles, We can say AC is perpendicular to BD (intersecting at point E - assume)
So, Quad. ABCD can be a rhombus!
Side Diagonal BD will then bisect i.e. BE = DE = $5\sqrt{3}$

In triangle AED:
Angle ADE = 30 with side opposite to it as $x$
Angle DAE = 60 with side opposite to it as $\sqrt{3}x$
Angle AED = 90 with side opposite to it as $2x$

Since DE = $5\sqrt{3}$
Then, AD = (2)(5) = 10

This is true for all four triangles. So, AB = BC = CD = AD = 10
Therefore the Perimeter = 40

Hence, option D