Finding the area of an orthic triangle (DEF) when given vertices of triangle ABC. Given the triangle ABC whose vertices are endpoints of the altitudes from $A$,$B$ and $C$ is called the orthic triangle. The triangle $ABC$ has vertices at $A=(2,4)$; $B=(8,5)$ and $C=(3,9)$. The altitude from $B$ to $AC$, meets AC at point $D=(2.42,6.12)$. Find the area if the orthic triangle.
To attempt this problem I decided to use the formula 
$$area = \frac{abc|cosAcosBcosC|}{2R}$$ where $R$ is the circumradius of the triangle $ABC$. I calculated the length of each sides by using the length equation and got $a=\sqrt41$, $b=\sqrt26$, and $c=\sqrt37$.
Next I used the cosine rule to get the angle at vertix $A$. Then used the sine rule to get angle at $B$ and subtracted these 2 angles from 180 to get the third angle. $A=69.23$, $B=48.12$ and $C=62.65$. 
To get the circumradius of the triangle $ABC$ I used $$R=\frac{abc}{4(Area)}$$ I found the area using heron's formula to be 14.5, putting all these values into the original equation I got the area of the orthic triangle to be 3.154. Is this correct? Is there an easier method? Thanks

I suggest to use
$$ \Delta_{orthic} = 2\Delta\left|\cos A\cos B\cos C\right|$$
and to compute $\Delta$ from the shoelace formula and the remaining part from the cosine theorem, such that we do not need to extract any square root. By the shoelace formula
$$ 2\Delta = |2\cdot 5+8\cdot 9+3\cdot 4-4\cdot 8-5\cdot 3-9\cdot 2|=29 $$
and by the cosine theorem
$$\left|\cos A\cos B\cos C\right|=\frac{(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^2+b^2+c^2)}{8a^2b^2 c^2} $$
where $a^2=41,b^2=26,c^2=37$ are given by the Pythagorean theorem. It follows that
$$\Delta_{orthic} = 29\cdot\frac{30\cdot 52\cdot 22 }{8\cdot 41\cdot 26\cdot 37}=\frac{4785}{1517}. $$
