smc

First realize that $△BCY\sim △DAX.$ Thus, because $CY:XA=2:3,$ we can say that $BY=2s$ and $DX=3s.$ From the Pythagorean Theorem, we have $A{B}^2=(2s{)}^2+4^2=4s^2+16$ and $C{D}^2=(3s{)}^2+3^2=9s^2+9.$ Because $AB=CD,$ from the problem statement, we have that $$4s^2+16=9s^2+9.$$ Solving, gives $s=\frac{\sqrt{7}}{\sqrt{5}}.$ To find the area of the trapezoid, we can compute the area of $△ABC$ and add it to the area of $△ACD.$ Thus, the area of the trapezoid is $\frac{1}{2}\cdot 2\cdot \frac{\sqrt{7}}{\sqrt{5}}\cdot 6+\left(\frac{1}{2}\cdot 3\cdot \frac{\sqrt{7}}{\sqrt{5}}\cdot 6\right)=\frac{15\sqrt{7}}{\sqrt{5}}=3\sqrt{35}.$ Thus, the answer is $\boxed{3\sqrt{35}}.$