In a triangle ABC,if r_1=3, r_2=4, r_3=6,then | vedclass

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(A) We know the formulas for the exradii of a triangle: $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$.Given $r_1=3, r

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$2 \sqrt{6}$

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(A) We know the formulas for the exradii of a triangle: $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$.Given $r_1=3, r_2=4, r_3=6$.Taking reciprocals: $\frac{1}{r_1} = \frac{s-a}{\Delta}, \frac{1}{r_2} = \frac{s-b}{\Delta}, \frac{1}{r_3} = \frac{s-c}{\Delta}$.Summing these: $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{3s-(a+b+c)}{\Delta} = \frac{3s-2s}{\Delta} = \frac{s}{\Delta} = \frac{1}{r}$.$\frac{1}{3} + \frac{1}{4} + \frac{1}{6} = \frac{4+3+2}{12} = \frac{9}{12} = \frac{3}{4} = \frac{1}{r} \implies r = \frac{4}{3}$.Also,$\Delta^2 = r r_1 r_2 r_3 = \frac{4}{3} 	imes 3 	imes 4 	imes 6 = 96 \implies \Delta = \sqrt{96} = 4\sqrt{6}$.Since $r_2 = \frac{\Delta}{s-b}$,we have $4 = \frac{4\sqrt{6}}{s-b} \implies s-b = \sqrt{6}$.Also $s = \frac{\Delta}{r} = \frac{4\sqrt{6}}{4/3} = 3\sqrt{6}$.Thus,$b = s - (s-b) = 3\sqrt{6} - \sqrt{6} = 2\sqrt{6}$.

In a triangle $ABC$,if $r_1=3, r_2=4, r_3=6$,then $b=$

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