Let the ellipse \( E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), \( a>b \) and \( E_2 : \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \), \( A<B \), have the 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: