5

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.

Brahmagupta
  • 4,204
  • 1
    in case you have the answers, all you have to do, is to check some triangles. just let $x=y=z=1$ , than you get Equilateral triangle with $a=\sqrt{2}$. than , if I see correctly , only one answer is valid. because the are of such triangle is $\frac{\sqrt{3}}{2}$ – d_e Aug 02 '15 at 07:54
  • 1
    HINT:

    $$A(x,0,0);B(0,y,0),C(0,0,z)$$

    See http://math.stackexchange.com/questions/128991/how-to-calculate-area-of-3d-triangle

     – lab bhattacharjee Aug 02 '15 at 07:57
  • @Brahmagupta . Hope Brahmagupta got fully clarified about Barahmagupta's formula. – Narasimham Aug 02 '15 at 08:05
  • @Narasimham,yes Sir,i did. – Brahmagupta Aug 02 '15 at 08:07

4 Answers4

5

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 $${\rm area\,}(\triangle)={1\over2}\bigl|{\bf a}\times{\bf b}\bigr|={1\over2}\sqrt{y^2z^2+z^2x^2+x^2y^2}\ .$$

Christian Blatter
  • 226,825
3

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 $S$.

DeepSea
  • 77,651
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

Harish Chandra Rajpoot
  • 37,450
2

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}

g.kov
  • 13,581