Question

If the sides of a triangle are $\frac{y}{z}+\frac{z}{x},\frac{z}{x}+\frac{x}{y}$ and $\frac{x}{y}+\frac{y}{z}$ then its area is

• A. xyz
• B. $\sqrt{\frac{x}{y}+\frac{y}{z}+\frac{z}{x}}$
• C. $\sqrt{xyz}$
• D. $\frac{x^2{y}^2{z}^2}{2}$

Answer 1

Answer: (B)

$\sqrt{\frac{x}{y}+\frac{y}{z}+\frac{z}{x}}$

Sides of triangle $\Rightarrow (\frac{y}{z}+\frac{z}{x},\frac{z}{x}+\frac{x}{y},\frac{x}{y}+\frac{y}{z})$

semiperimeter $=s=\frac{sumofside}{2}$

$=\frac{(\frac{y}{z}+\frac{z}{x})+(\frac{z}{x}+\frac{x}{y})+(\frac{x}{y}+\frac{y}{z})}{2}$

$S=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$

From Heron's Formula, area of triangle $\Delta =\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-\frac{y}{z}-\frac{z}{x})(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-\frac{z}{x}-\frac{x}{y})(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-\frac{x}{y}-\frac{y}{z})}$

$=\sqrt{\left(\frac{x}{y}\right)\left(\frac{y}{z}\right)\left(\frac{z}{x}\right)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})}$

area of triangle $=\sqrt{(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})}$