Coordinate Geometry:-)

Let the isosceles trapezoid (which is symmetric about the $x-$ axis) have coordinates: ( $x_{A}, \pm 2\sqrt{ax_{A}}); (x_{B}, \pm 2\sqrt{ax_{B}})$ for $0 < x_{A} < 1 < x_{B}.$ We now derive the formulae for the area and the twin-diagonal length:

$A = \frac{x_{B} - x_{A}}{2} \cdot (4\sqrt{ax_{A}} + 4\sqrt{ax_{B}});$ (i)

$\frac{25}{4} = \sqrt{(x_{A}-x_{B})^2 + (2\sqrt{ax_{A}} + 2\sqrt{ax_{B}})^2}$ (ii).

Upon squaring (ii) we obtain:

$(2\sqrt{ax_{A}} + 2\sqrt{ax_{B}})^2 = (\frac{25}{4})^2 - (x_{A}-x_{B})^2 \Rightarrow \boxed{2\sqrt{ax_{A}} + 2\sqrt{ax_{B}} = \sqrt{(\frac{25}{4})^2 - (x_{A}-x_{B})^2}}$

which when substituted into (i) produces:

$A = \frac{x_{B} - x_{A}}{2} \cdot (4\sqrt{ax_{A}} + 4\sqrt{ax_{B}}) = (x_{B} - x_{A}) \cdot (2\sqrt{ax_{A}} + 2\sqrt{ax_{B}}) = (x_{B}-x_{A}) \cdot \sqrt{\frac{25^2 - 4^2(x_{A}-x_{B})^2}{16}} = \frac{x_{B}-x_{A}}{4} \cdot \sqrt{25^2 - (4(x_{A}-x_{B}))^2}$ (iii).

We next find that each of our answer choices is rational, and this requires the radicand to be a perfect-square integer. Drawing upon the Pythagorean triplets $(15,20,25)$ and $(7,24,25)$ gives us the prospective quantities:

$x_{B}-x_{A} = 5$ or $6$

and respective areas of $A = \frac{75}{4}, \frac{21}{2}$ . Thus, Choice D is our desired answer.