Q73 / 160 · AP EAMCET 2008 · Mathematics

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

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

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, $$ \begin{aligned} \frac{a}{b}+\frac{b}{c}+\frac{c}{a} & =\frac{5}{4}+\frac{4}{3}+\frac{3}{5} \\ & =\frac{75+80+36}{60}=\frac{191}{60} \end{aligned} $$

Hence, the correct answer is option: 'D'.

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