In triangle $A B C$, if $r_1=36, r_2=18$ and $r_3=12$, then $a+b=$

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Aicharya · Accepted Answer

It is given in a $\triangle A B C$ $\begin{aligned} & r_1=\frac{\Delta}{s-a}=36, r_2=\frac{\Delta}{s-b}=18 \text { and } r_3=\frac{\Delta}{s-c}=12 \ \Rightarrow & \frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{36}+\frac{1}{18}+\frac{1}{12}=\frac{1+2+3}{36}=\frac{1}{6} \ \Rightarrow & \frac{s-a+s-b+s-c}{\Delta}=\frac{1}{6} \ \Rightarrow & \frac{s}{\Delta}=\frac{1}{6} \Rightarrow \Delta=6 s \end{aligned}$ So, $r_1=\frac{\Delta}{s-a}=\frac{6 s}{s-a}=36 \Rightarrow s=6 s-6 a \Rightarrow 6 a=5 s$ $r_2=\frac{\Delta}{s-b}=\frac{6 s}{s-b}=18 \Rightarrow s=3 s-3 b \Rightarrow 3 b=2 s$ and $r_3=\frac{\Delta}{s-c}=\frac{6 s}{s-c}=12 \Rightarrow s=2 s-2 c \Rightarrow 2 c=s$ So, $\Delta^2=36 s^2$ $\begin{array}{lrl} \Rightarrow & s(s-a)(s-b)(s-c)=36 s^2 \ \Rightarrow & \left(s-\frac{5 s}{6}\right)\left(s-\frac{2 s}{3}\right)\left(s-\frac{s}{2}\right)=36 \mathrm{~s} \ \Rightarrow & s^3\left(\frac{1}{6}\right)\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)=36 s \ \Rightarrow & s^2=(36)^2 \Rightarrow s=36 \ \therefore & a & =30 \text { and } b=24 \end{array}$ So, $a+b=30+24=54$ Hence, option (d) is correct.

In triangle \(A B C\), if \(r_1=36, r_2=18\) and \(r_3=12\), then \(a+b=\)

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