In a triangle, if $r_1=2 r_2=3 r_3$, then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ is equal to
In a triangle, if $r_1=2 r_2=3 r_3$, then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ is equal to
- $\frac{75}{60}$
- $\frac{155}{60}$
- $\frac{176}{60}$
- $\frac{191}{60}$
Solution
Given that, $r_1=2 r_2=3 r_3$
$
\therefore \quad \frac{\Delta}{s-a}=\frac{2 \Delta}{s-b}=\frac{3 \Delta}{s-c}=\frac{\Delta}{k}
$
Then, $s-a=k, s-b=2 k, s-c=3 k$
$
\begin{array}{rlrl}
\Rightarrow & 3 s-(a+b+c) & =6 k \Rightarrow s=6 k \\
\therefore & & \frac{a}{5}=\frac{b}{4} & =\frac{c}{3}=k
\end{array}
$
Now,
$
\begin{aligned}
\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}
\end{aligned}
$
Asked in: AP EAMCET 2008
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