I know the answer not the solution!

A circle with centre $O$ is circumscribed about the $\triangle ABC$ with $\angle A$ as an obtuse angle. The radius $AO$ forms an angle $30$ with the altitude $AH$ on $BC$ . The extension of angle bisector of $\angle A$ meets $BC$ at $F$ and the circumference of the circle at $L$ and the radius $AO$ intersects $BC$ at point $E$ . Compute the area of the quadrilateral $FEOL$ if it is known that $AL=4\sqrt{2}$ $\text{cm}$ and $AH=\sqrt{2\sqrt{3}}$ $\text{ cm}$ .

If your answer comes as $a(b-\sqrt{c})$ $\text{ cm}^{2}$ submit it as $a+b+c$ .

Note : All angles are in degree.