Question

In a triangle, if $r_1=2r_2=3r_3$, then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ is equal to

• A. $\frac{75}{60}$
• B. $\frac{155}{60}$
• C. $\frac{176}{60}$
• D. $\frac{191}{60}$

Answer 1

Answer: (D)

$\frac{191}{60}$

Given that, $r_1=2r_2=3r_3$
$\therefore \frac{\Delta }{s-a}=\frac{2\Delta }{s-b}=\frac{3\Delta }{s-c}=\frac{\Delta }{k}$ (say)
Then, $s-a=k,s-b=2k,s-c=3k$
$\Rightarrow 3s-(a+b+c)=6k$

$\Rightarrow s=6k$
Therefore, $\frac{a}{5}=\frac{b}{4}=\frac{c}{3}=k$
Now, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{5}{4}+\frac{4}{3}+\frac{3}{5}$

$=\frac{75+80+36}{60}=\frac{191}{60}$