Area of a triangle with sides $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$,$\sqrt{z^2+x^2}$ Sides of a triangle ABC are $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$ and $\sqrt{z^2+x^2}$ where x,y,z are non-zero real numbers,then area of triangle ABC is
(A)$\frac{1}{2}\sqrt{x^2y^2+y^2z^2+z^2x^2}$
(B)$\frac{1}{2}(x^2+y^2+z^2)$
(C)$\frac{1}{2}(xy+yz+zx)$
(D)$\frac{1}{2}(x+y+z)\sqrt{x^2+y^2+z^2}$
I tried applying Heron's formula but calculations are very messy and simplification is difficult.I could not think of any other method to find this area.Can someone assist me in solving this problem.

For such form of the side lengths, the most convenient 
would be a variation of the 
Heron's formula
for the area:
\begin{align}
S&=\tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}
\\
S&=\tfrac14\sqrt{
4(x^2+y^2)(y^2+z^2)-
(x^2+y^2+y^2+z^2-z^2-x^2)^2
}
\\
&=\tfrac14\sqrt{4(x^2+y^2)(y^2+z^2)-4y^4}
\\
&=\tfrac12\sqrt{x^2 y^2+y^2 z^2+z^2 x^2}.
\end{align}
