What is the value of $a$ where $√a$ is area of a trapezoid which touches the circle with center $O$ (diameter is 2)? 

The sides $AB, BC, CD$ of trapezoid $ABCD$ touches the circle with
  center $O$ and they are equal. $AD$, goes through the point O. If
  diameter is 2, then the area of the trapezoid is $√a$ . What is
  the value of a?

Source: Bangladesh Math Olympiad 2016 Junior Catagory
I tried but I could not find any possible solution. How am I supposed to get $BC + AD$?
 A: Warning: might be a little convoluted, uses a bit of assumed knowledge, and is probably not the most elegant/efficient solution. I've triple-checked the method and arithmetic though so unless there's a fundamental flaw with my solution, I think this should be correct.

We make a duplicate trapezoid of $ABCD$, with diameter $AD$, which basically inscribes the circle in a regular hexagon:

Since the duplicate trapezoid has the same area, the area of the hexagon is $2\sqrt a$.
The area of a regular hexagon can be shown to be given by $\frac{3 \sqrt 3}{2} s^2$, where $s$ is the length of a side of the hexagon. Thus,
$$2 \sqrt a = \frac{3 \sqrt 3}{2} s^2 \implies a = \frac{27}{16} s^4$$
The area of a regular polygon can also be shown (link above) to be given by $xp/2$, for perimeter $p$ and apothem $x$. Here, $p = 6s$, and $x = 1$, the radius of the circle, clear by the construction above and that the diameter is $2$. Thus, the area is given by $(6s)(1)/2 = 3s$.
Thus,
$$2 \sqrt a = 3s \implies a = \frac{9}{4}s^2$$
Thus,
$$a = \frac{27}{16}s^4 = \frac{9}{4}s^2$$
We subtract the right fraction from the left, and multiply through by $16$ as we begin solving for $s$. We get the equation
$$27s^4 - 36s^2 = 0$$
Let $u = s^2$ for ease of use, then the above equation becomes
$$27u^2 - 36u = 0 \implies 9u(3u - 4) = 0 \implies u = 0, u = \frac{4}{3}$$
$u=0$ is obviously not what we want, but $u=4/3$ is fine. Then, going back through our substitutions,
$$u=\frac{4}{3} \implies s^2 = \frac{4}{3} \implies s = \frac{2}{\sqrt 3} \implies 2 \sqrt a = 3 \cdot \frac{2}{\sqrt 3} \implies a = 3$$
A: Hint. 
Once you showed that $ABO$, $OBC$ and $COD$ are equilater triangles, note that the altitude of such triangles is equal to $r=1$. By Pythagorean theorem on half of the triangle determine the side, i.e. $\frac{2\sqrt{3}}{3}$. The total area of the trapezoid is equal to three times the area of the triangle, i.e. $\sqrt{3}$
