In a triangle ABC,if r_1=36, r_2=18 and r_3=1 | vedclass

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(D) Given,$r_1=36, r_2=18, r_3=12$. We know that in a $	riangle ABC$,the relation between the inradius $r$ and exradii $r_1, r_2, r_3$ is $\frac{1}{r

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$36$

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(D) Given,$r_1=36, r_2=18, r_3=12$. We know that in a $	riangle ABC$,the relation between the inradius $r$ and exradii $r_1, r_2, r_3$ is $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$. Substituting the values: $\frac{1}{r} = \frac{1}{36} + \frac{1}{18} + \frac{1}{12} = \frac{1+2+3}{36} = \frac{6}{36} = \frac{1}{6}$. Thus,$r=6$. We also know that $\Delta^2 = r \cdot r_1 \cdot r_2 \cdot r_3$. $\Delta^2 = 6 	imes 36 	imes 18 	imes 12 = 6 	imes 6^2 	imes (6 	imes 3) 	imes (6 	imes 2) = 6^4 	imes 36 = 6^4 	imes 6^2 = 6^6$. Therefore,$\Delta = 6^3 = 216$. Since $r = \frac{\Delta}{s}$,we have $s = \frac{\Delta}{r} = \frac{216}{6} = 36$.

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

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