Question

If $(\frac{2}{5}\frac{y}{z}+\frac{2}{5}\frac{z}{x})$, $(\frac{2}{5}\frac{z}{x}+\frac{2}{5}\frac{x}{y})$ , $(\frac{2}{5}\frac{x}{y}+\frac{2}{5}\frac{y}{z})$ are the sides of a triangle then its area is

• A. $\frac{4}{25}\sqrt{\frac{x}{y}+\frac{y}{z}+\frac{z}{x}}$
• B. $xyz$
• C. $1$
• D. $x+y+z$

Answer 1

The correct option is A $\frac{4}{25}\sqrt{\frac{x}{y}+\frac{y}{z}+\frac{z}{x}}$

Let $a,b,c$ be the sides of given triangle

$s$ be semiperimeter of the triangle

Given, $a=\frac{2}{5}\times \frac{y}{z}+\frac{2}{5}\times \frac{z}{x}$

$b=\frac{2}{5}\times \frac{z}{x}+\frac{2}{5}\times \frac{x}{y}$

$c=\frac{2}{5}\times \frac{x}{y}+\frac{2}{5}\times \frac{y}{z}$

Therefore, $s=\frac{2}{5}\times \frac{x}{y}+\frac{2}{5}\times \frac{y}{z}+\frac{2}{5}\times \frac{z}{x}$

$(s-a)=\frac{2}{5}\times \frac{x}{y}$

$(s-b)=\frac{2}{5}\times \frac{y}{z}$

$(s-c)=\frac{2}{5}\times \frac{z}{x}$

Since, $\text{area of triangle},\Delta =\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

Therefore, $\Delta =\sqrt{\frac{8}{125}(\frac{2}{5}\times \frac{x}{y}+\frac{2}{5}\times \frac{y}{z}+\frac{2}{5}\times \frac{z}{x})}$

$\Rightarrow$ $\Delta =\left(\frac{4}{25}\right)\sqrt{(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})}$