Is this figure possible to achieve?
$AD > BC$; $\angle BCD$ = $\angle BAD$ = $90^{\circ}$ .
And this is supposed to be a closed quadrilateral.
I'm confused as to whether this is viable - since $AD > BC$, and the angles are $90^{\circ}$, how would this figure close?
The figure seems perfectly alright. The confusion in your mind may be because you are considering the angle at D to be $90^{\circ}$. Imagine it to be an acute angle and then it would be fine. Note that one of the hypotenuse BD is $\sqrt{34}$ by pythagorus theorem. And hence the remaining side is of length $4$. Try to draw the figure with $D$ as an acute angle and your confusion would be clear.