Right tangential trapezoid 
In a right tangential trapezoid $ABCD$ $(AB\parallel CD)$ and $AD\perp AB$ the incircle is $k(O).$ Find the area of the trapezoid if $CO=6$ and $BO=8.$

The triangle $BOC$ is a right triangle and by the Pythagorean theorem we get $BC=10.$ I don't see how this and $AB+CD=AD+BC$ is enough to find the area. Can you help me?  
 A: Hint: 
Let $r$ be the incircle radius. Since the angle $BOC$ is right, we have:
$$r=\frac{OB\cdot OC}{BC}=\frac{24}5.$$

 $$[ABCD]=r(AD+BC)=r(2r+BC)=\frac{24}{5}\left(\frac{48}5+10\right).$$

A: You need to do a little more geometry to get there.
Draw the line through $O$ parallel to $\overline{AD}$.  this intersects  $\overline {AB}$ at $X$ and $\overline{CD}$ at $Y$, both of these points also being tangent points on the respective trapezoid sides.
Next draw $\overline{OB}$ and $\overline{OC}$.  These bisect the vertex angles at $B$ and $C$, causing all three of the triangles $BOC, OXB, BYO$ to be similar.  From that similarity infer that the radius of the circle is $(OB)(OC)/BC=24/5$.  The height of the trapezoid is twice that.
That height is also the length of the side $\overline{AD}$ which, in the right trapezoid, is perpendicular to the bases.  So now you have $BC+AD$ and therefore $AB+CD$, where the latter is the sum of the two bases.  Multiply that by half the height which was figured out earlier and you have the required area.
A: If BOC is right triangle we have:
$r=\frac {6\times8}{10}=4.8$
Now extend CO to intersect AB in D, we have:
$S_{BCD}=8\times 6=48$
If you draw a line parallel with AD crossing O it intersects AB and DC on E and F respectively and we have:
$S_{AEF}=4.8\times 9.6=46.08$
The area of triangles ODE and OFC are equal, so the area of ABCD is:
$S_{ABCD}=S_{BDC}+S_{AEFD}=48+46.08=94.08$
