In a triangle ABC,if r_1=12, r_2=18 and r_3=3 | vedclass

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(C) Given,in $	riangle ABC$,$r_1=12, r_2=18, r_3=36$. We know that $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$. Substituting the va

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$24$

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(C) Given,in $	riangle ABC$,$r_1=12, r_2=18, r_3=36$. We know that $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$. Substituting the values: $\frac{1}{r} = \frac{1}{12} + \frac{1}{18} + \frac{1}{36} = \frac{3+2+1}{36} = \frac{6}{36} = \frac{1}{6}$. Thus,$r = 6$. Also,$\Delta^2 = r_1 r_2 r_3 r = 12 	imes 18 	imes 36 	imes 6 = 46656$. So,$\Delta = \sqrt{46656} = 216$. Since $r = \frac{\Delta}{s}$,we have $s = \frac{\Delta}{r} = \frac{216}{6} = 36$. We know $r_2 = \frac{\Delta}{s-b}$. Substituting the values: $18 = \frac{216}{36-b}$. $36-b = \frac{216}{18} = 12$. $b = 36 - 12 = 24$.

In a $\triangle ABC$,if $r_1=12, r_2=18$ and $r_3=36$,then $b=$

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