If in a triangle ABC,r_1 = 2r_2 = 3r_3,then t | vedclass

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(B) Given $r_1 = 2r_2 = 3r_3 = \lambda$ (say).We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.So,$\f

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$3b$

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(B) Given $r_1 = 2r_2 = 3r_3 = \lambda$ (say).We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.So,$\frac{\Delta}{s-a} = \lambda$,$\frac{\Delta}{s-b} = \frac{\lambda}{2}$,and $\frac{\Delta}{s-c} = \frac{\lambda}{3}$.This gives $s-a = \frac{\Delta}{\lambda}$,$s-b = \frac{2\Delta}{\lambda}$,and $s-c = \frac{3\Delta}{\lambda}$.Adding these,$(s-a) + (s-b) + (s-c) = \frac{\Delta}{\lambda} (1 + 2 + 3) = \frac{6\Delta}{\lambda}$.$3s - (a+b+c) = \frac{6\Delta}{\lambda}$.Since $a+b+c = 2s$,we have $3s - 2s = s = \frac{6\Delta}{\lambda}$.Thus,$s-a = \frac{s}{6}$,$s-b = \frac{2s}{6} = \frac{s}{3}$,and $s-c = \frac{3s}{6} = \frac{s}{2}$.Solving for sides: $a = s - \frac{s}{6} = \frac{5s}{6}$,$b = s - \frac{s}{3} = \frac{2s}{3}$,$c = s - \frac{s}{2} = \frac{s}{2}$.Perimeter $P = a+b+c = \frac{5s}{6} + \frac{4s}{6} + \frac{3s}{6} = \frac{12s}{6} = 2s$.Also,$b = \frac{2s}{3} \implies 3b = 2s = P$. Therefore,the perimeter is $3b$.

If in a $\triangle ABC$,$r_1 = 2r_2 = 3r_3$,then the perimeter of the triangle is equal to

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