If in a triangle ABC,r_1=2,r_2=3 and r_3=6,th | vedclass

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(D) Given,$r_1=2$,$r_2=3$,and $r_3=6$. We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$,where $r$ is the inradius. $\frac{1}

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$3$

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(D) Given,$r_1=2$,$r_2=3$,and $r_3=6$. We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$,where $r$ is the inradius. $\frac{1}{r} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1$. So,$r = 1$. Also,$\Delta = \sqrt{r r_1 r_2 r_3} = \sqrt{1 	imes 2 	imes 3 	imes 6} = \sqrt{36} = 6$. Since $r_1 = \frac{\Delta}{s-a}$,we have $2 = \frac{6}{s-a}$,which implies $s-a = 3$. Also,$s = \frac{\Delta}{r} = \frac{6}{1} = 6$. Substituting $s=6$ into $s-a=3$,we get $6-a=3$,so $a=3$.

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

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