If in a $\triangle A B C, r_1=2, r_2=3$ and $r_3=6$, then $a$ equals to

If in a $\triangle A B C, r_1=2, r_2=3$ and $r_3=6$, then $a$ equals to
  1. $4$
  2. $1$
  3. $2$
  4. $3$

Solution

Given, $r_1=2, r_2=3$ and $r_3=6$ $ \begin{aligned} & \because \quad r_1=\frac{\Delta}{s-a} \Rightarrow 2=\frac{\Delta}{s-a} \\ & \Rightarrow \frac{s-a}{\Delta}=\frac{1}{2} \\ & r_2=\frac{\Delta}{s-b} \Rightarrow 3=\frac{\Delta}{s-b} \\ & \Rightarrow \quad \frac{s-b}{\Delta}=\frac{1}{3} \\ & \text { and } \\ & r_3=\frac{\Delta}{s-c} \Rightarrow 6=\frac{\Delta}{s-C} \\ & \Rightarrow \quad \frac{s-c}{4}=\frac{1}{6} \\ & \end{aligned} $ On adding Eqs. (i), (ii) and (iii), we get $ \begin{aligned} & \frac{s-a}{\Delta}+\frac{(s-b)}{4}+\frac{s-c}{\Delta}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6} \\ \Rightarrow & \frac{3 s-(a+b+c)}{4}=\frac{3+2+1}{6} \\ \Rightarrow & \frac{3 s-2 s}{4}=\frac{6}{6}=0 \\ \Rightarrow & \frac{s}{\Delta}=1 \\ \because & s^2=r_1 r_2+r_2 r_3+r_3 r_1 \end{aligned} $ $ \begin{aligned} & =2 \times 3+3 \times 6+6 \times 2 \\ & =6+18+12=36 \\ \Rightarrow \quad s^2 & =36 \Rightarrow s=6 \end{aligned} $ From Eq. (i), we get $ \begin{aligned} \frac{6}{\Delta} & =1 \\ \text { Now, } \quad \quad \quad \quad \quad r_1 & =\frac{\Delta}{s-a} \\ 2 & =\frac{6}{6-a} \\ \Rightarrow \quad 6-a & =3 \Rightarrow a=3 \end{aligned} $

Asked in: AP EAMCET 2014

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