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.

hint: Let $a = \sqrt{x^2+y^2}, b = \sqrt{y^2+z^2}, c = \sqrt{z^2+x^2} \to a^2 = x^2+y^2, b^2 = y^2+z^2, c^2= z^2+x^2 $. Use this and Cosine Law to find $\cos^2 A$, then $\sin^2 A$, and use $S^2 = \dfrac{b^2c^2\sin^2 A}{4}$, to find $S^2$ and then take square-root to get back $S$. 
