Right tangential trapezoid

Question

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?

Answer 1

Hint:

Let $r$ be the incircle radius. Since the angle $BOC$ is right, we have: $$r=\frac{OB\cdot OC}{BC}=\frac{24}5.$$

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

Answer 2

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.

Answer 3

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:

$S_{ABCD}=S_{BDC}+S_{AEFD}=48+46.08=94.08$