What is the area of the triangle having side lengths $\frac{y}{z} + \frac{z}{x},\frac{z}{x} + \frac{x}{y},\frac{x}{y} + \frac{y}{z}?$

Question

TuteeHUB · Accepted Answer

$\sqrt {\frac{{xy + yz + zx}}{{xyz}}} $

TuteeHUB · Answer

$\frac{{{{\left( {x + y + z} \right)}^2}}}{{xyz}}$

TuteeHUB · Answer

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

TuteeHUB · Answer

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

1.	What is the area of the triangle having side lengths \(\frac{y}{z} + \frac{z}{x},\frac{z}{x} + \frac{x}{y},\frac{x}{y} + \frac{y}{z}?\)
A.	\(\frac{{{{\left( {x + y + z} \right)}^2}}}{{xyz}}\)
B.	\(\frac{{\sqrt {xyz} }}{{x + y + z}}\)
C.	\(\sqrt {\frac{x}{y} + \frac{y}{z} + \frac{z}{x}} \)
D.	\(\sqrt {\frac{{xy + yz + zx}}{{xyz}}} \)
Answer» D. \(\sqrt {\frac{{xy + yz + zx}}{{xyz}}} \)

What is the area of the triangle having side lengths \(\frac{y}{z} + \frac{z}{x},\frac{z}{x} + \frac{x}{y},\frac{x}{y} + \frac{y}{z}?\)