Area of a triangle with sides $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$,$\sqrt{z^2+x^2}$

Question

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.

Answer 1

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 $S$.

Answer 2

Use cosine rule to find say $\angle C$ then use formula of area as follows

Area of $\triangle ABC$ $$=\frac{1}{2}(a)(b)\sin C=\frac{1}{2}(a)(b)\sqrt{1-(\cos C)^2}$$ $$=\frac{1}{2}(\sqrt{x^2+y^2})(\sqrt{y^2+z^2})\sqrt{1-\left(\frac{(\sqrt{x^2+y^2})^2+\sqrt{y^2+z^2})^2-(\sqrt{x^2+z^2})^2}{2\sqrt{x^2+y^2}\sqrt{y^2+z^2}}\right)^2}$$ $$=\frac{1}{2}(\sqrt{x^2+y^2})(\sqrt{y^2+z^2})\sqrt{\frac{4(x^2+y^2)(y^2+z^2)-(x^2+y^2+y^2+z^2-x^2-z^2)^2}{4(x^2+y^2)(y^2+z^2)}}$$

$$=\frac{1}{2}\frac{(\sqrt{x^2+y^2})(\sqrt{y^2+z^2})}{2(\sqrt{x^2+y^2})(\sqrt{y^2+z^2})}\sqrt{4x^2y^2+4y^4+4z^2x^2+4y^2z^2-(2y^2)^2}$$ $$=\frac{1}{4}\sqrt{4(x^2y^2+y^2z^2+z^2x^2)}$$ $$=\frac{1}{2}\sqrt{x^2y^2+y^2z^2+z^2x^2}$$ Option (A) is correct

Answer 3

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}

Answer 4

There is such a triangle $\triangle\subset{\mathbb R}^3$ with vertices $${\bf 0}=(0,0,0), \quad{\bf a}:=(x,y,0), \quad {\bf b}:=(0,y,z)\ .$$ Its area is given by $${\rm area\,}(\triangle)={1\over2}\bigl|{\bf a}\times{\bf b}\bigr|={1\over2}\sqrt{y^2z^2+z^2x^2+x^2y^2}\ .$$