Isosceles trapezoid with inscribed circle The area an isosceles trapezoid is equal to $S$, and the height is equal to the half of one of the non-parallel sides. If a circle can be inscribed in the trapezoid, find, with the proof, the radius of the inscribed circle. Express your answer in terms of $S$ only.
I labeled the trapezoid $ABCD$ starting lower left corner going clockwise. The area $S$ is equal to $h\times\left({a+b\over 2}\right)$. So $S=\left({AD+BC\over 2}\right)\times\left({AB\over 2}\right)=\left({AB\times(AD+BC)\over 4}\right)$. I know intuitively that because the circle is inscribed and the tangents are parallel, the two perpendicular radii form the diameter, but I don't know how to prove that (I need to). From there its the same as the height I would guess, unsure how to proceed. 


Let the radius of the circle be $r$; then the height of the isosceles trapezoid is $2r$, and the length of a lateral side would be $4r$.
The four right triangles with $OB$ and $OC$ as hypotenuses are congruent.  The four right triangles with $OA$ and $OD$ as hypotenuses are also congruent.  Therefore the lengths marked $x$ are all the same, as are those marked $y$.
The area of a trapezoid is $$\frac{1}{2}h(a+b)$$
Therefore $$S = \frac{1}{2}h(a+b)$$
$$S = \frac{1}{2}(2r)(2x + 2y)$$
$$S = 2r(x + y)$$
But we know $x + y = 4r$.  Therefore,
$$S = 2r(4r)$$
$$\therefore r = \sqrt{\frac{S}{8}} = \frac{\sqrt{2}}{4}\sqrt{S} \approx 0.3536 \sqrt{S}$$
