Area of Triangle........

[man111]

The Sides of a [tex]\triangle[/tex] is Given by [tex]\sqrt{b^2+c^2}, \sqrt{c^2+a^2}, \sqrt{a^2+b^2}[/tex]. Then Its Area is

[ins-]

Use the following link:
http://en.wikipedia.org/wiki/Heron%27s_formula
Substiture and you will get the answer.

[mkmarinov]

Let the angle opposite [tex]\sqrt{a^2+b^2}[/tex] be [tex]\varphi[/tex].
[tex]a^2+b^2=a^2+c^2+b^2+c^2-2\sqrt{a^2+c^2}\sqrt{b^2+c^2}cos\varphi[/tex]
[tex]cos\varphi=\frac{2c^2}{2\sqrt{a^2+c^2}\sqrt{b^2+c^2}}=\frac{c^2}{\sqrt{a^2+c^2}\sqrt{b^2+c^2}}[/tex]
[tex]sin\varphi = \sqrt{1-cos^2\varphi}=\sqrt{1-\frac{c^4}{(a^2+c^2)(b^2+c^2)}}=\sqrt{\frac{a^2b^2+c^2(a^2+b^2)}{(a^2+c^2)(b^2+c^2)}}[/tex]
[tex]S=\frac{1}{2}\sqrt{a^2+c^2}\sqrt{b^2+c^2}sin\varphi = \frac{1}{2}\sqrt{a^2b^2+a^2c^2+b^2c^2}[/tex]