ABCD is a paper card where BA perpendicular to DA .It is given that AB=BC=CD=6cm and AD=8cm

I think mine is right Chris.  

You are right I did describe ABD staying in the horizontal plane - not BDC as described - but this makes no difference to the volume.  I did it because it made the calculations easier.

I have decided to stick by my answer.  For now anyway.  (I always leave my options open LOL )

Maybe Gino or Alan could arbitrate here ?

This is a trianglular pyramid

So volume = 1/3 * area of base * perpendicular height

base = 1/2*6*8=24cm squared

 $$height=\sqrt{11}*sin\theta$$

$$\\v=\frac{1}{3}*24*\sqrt{11}*sin\theta\\\\
v=8\sqrt{11}sin\theta\\\\$$

Volume is greatest when sin theta is greatest ie theta=90degrees

$$$Max volume $= 8\;\sqrt{11}\; cm^3$$

That is what I think anyway.   

In my diagram the right angled vertice is A The going clockwise B is at the top then C then D

I don't believe this might be correct, Melody.......your drawing has "A" at the bottom left, and, in order clockwise from this point, we have "B," "C" and "D."  And the problem states that "B," "C" and "D" lie on the horizontal plane. So this triangle must form the "base." And, by Heron's formula, its area is

A = √(s(s-BD)(s-BC)(s-CD))  where s = [10 + 6 + 6]/ 2   = 11 

So we have

A = √((11)(1)(5)(5)) = √(275) = 16.583 units  (???)

Actually....I'm not sure we have anything, here......If BCD lies on the horizontal plane, DAB is just some triangular "flap" that shares a common side with BCD.... ("A" doesn't touch the plane, according to the problem)

Maybe I'm seeing this incorrectly, or interpreting it incorrectly???