The following figure is given with the lengths of the sides and the diagonal:
We can circumscribe a circle around the quadrilateral and inscribe a circle in it. How can the area of the quadrilateral be found?
I do not understand what to do with these circles. How are they helping me?
From the law of cosines:
$$AC^2=AB^2+BC^2-2AB\cdot BC \cos \angle B$$
$$(10\sqrt7)^2=10^2+20^2-2\cdot10\cdot20\cos\angle B$$
$$700=100+400-400\cos\angle B$$
$$\cos\angle B=-\frac12$$
$$\angle B=120^\circ$$
The fact that you can draw a circumscribed circle means that quadrilateral $ABCD$ is cyclic. In a cyclic quadrilateral the sum of the opposite angles is $180^\circ$. So we have:
$$\angle D=180^\circ-\angle B=60^\circ$$
The fact that you can inscribe the circle in a quadruilateral means that the sum of opposite sides must be equal. Therefore:
$$AB+CD=BC+DA$$
$$10+CD=20+DA$$
Introduce $CD=x$. It means that $DA=x-10$. Now apply cosine theorem to triangle ACD:
$$CD^2+DA^2-2CD\cdot DA\cos\angle D=AC^2$$
$$x^2+(x-10)^2-2x(x-10)\cos60^\circ=(10\sqrt7)^2$$
$$x^2+(x-10)^2-x(x-10)=700$$
$$x^2-10x-600=0$$
$$x_{1,2}=\frac{10\pm\sqrt{10^2+4\cdot 600}}{2}=\frac{10\pm50}{2}$$
If you discard negative solution, you get:
$$x=CD=30$$
$$DA=x-10=20$$
Finding the area is simple:
$$A_{ABC}=\frac12 AB\cdot BC\sin\angle B=\frac12 10\cdot20\sin120^\circ=50\sqrt 3$$ $$A_{ACD}=\frac12 CD\cdot DA\sin\angle D=\frac12 30\cdot20\sin60^\circ=150\sqrt 3$$
...so the total area is:
$$A=A_{ABC}+A_{ACD}=200\sqrt3$$