Let the ellipse ${{\rm{E}}_1}:\frac{{{x^2}}}{{{{\rm{a}}^2}}} + \frac{{{y^2}}}{{\;{{\rm{b}}^2}}} = 1,{\rm{a}} > {\rm{b}}$ and ${{\rm{E}}_2}:\frac{{{x^2}}}{{\;{{\rm{A}}^2}}} + \frac{{{y^2}}}{{\;{{\rm{B}}^2}}} = 1,\;{\rm{A}} < {\rm{B}}$ have same eccentricity $\frac{1}{{\sqrt 3 }}$ . Let the product of their lengths of latus rectums be $\frac{{32}}{{\sqrt 3 }}$, and the distance between the foci of ${E_1}$ be 4. If ${E_1}$ and ${E_2}$ meet at A,B,C and D, then the area of the quadrilateral ABCD equals :