Given Orthocenter of a triangle and length of three segments, need to find area. Let, $O$ be an orthocenter of the triangle $\triangle ABC$, where $AO = 7, AB = 9, CO = 4$. Find the combined area of the triangles $\triangle AOC$ and $\triangle AOB$. Options given are $7\sqrt{3}, 14\sqrt{3}, 21\sqrt{3}, 28\sqrt{3}$.
I started off by assigning $AE =x$. Observing $\Delta AOE$, I can say $OE = \sqrt{49-x^2}$.
Then, $BE = 9-x$ and $OB = \sqrt{130-18x}$. Furthermore, $BC$ would be $\sqrt{146+8\sqrt{49-x^2}}$. But not sure if I'm going in right direction.

$$\triangle=[AOB]+[AOC]=\frac{1}{2}\cdot BC\cdot AO$$
So we need to find $BC=a$
We use the fact that perpendicular bisector from circumcenter to any side is half the length of segment joining the opposite vertex to the orthocenter.
In the diagram below, not to scale, we see that

$$\left( \frac{c}{2} \right)^2 + \left( \frac{OC}{2} \right)^2 = R^2 = \left( \frac{a}{2} \right)^2 +\left( \frac{AO}{2} \right)^2 $$
$$\Rightarrow a^2 = 4^2+9^2-7^2=48$$
$$\therefore \triangle=14\sqrt{3}$$
