In triangle ABC,frac{r_2(r_1+r_3)}{sqrt{r_1 r | vedclass

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(B) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$ and $\Delta^2 = s(s-a)(s-b)(s-c)$.Substituting these

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$b$

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(B) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$ and $\Delta^2 = s(s-a)(s-b)(s-c)$.Substituting these into the expression:$\frac{r_2(r_1+r_3)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}} = \frac{\frac{\Delta}{s-b}(\frac{\Delta}{s-a} + \frac{\Delta}{s-c})}{\sqrt{\frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}}}$$= \frac{\frac{\Delta^2}{s-b} \cdot \frac{s-c+s-a}{(s-a)(s-c)}}{\Delta \sqrt{\frac{s-c+s-a+s-b}{(s-a)(s-b)(s-c)}}}$$= \frac{\Delta \cdot b}{(s-a)(s-b)(s-c)} \cdot \sqrt{\frac{(s-a)(s-b)(s-c)}{3s-(a+b+c)}}$Since $a+b+c = 2s$,the denominator inside the square root is $3s-2s = s$.$= \frac{\Delta \cdot b}{(s-a)(s-b)(s-c)} \cdot \frac{\sqrt{(s-a)(s-b)(s-c)}}{\sqrt{s}}$$= \frac{\Delta \cdot b}{\sqrt{s(s-a)(s-b)(s-c)}} = \frac{\Delta \cdot b}{\Delta} = b$.

In $\triangle ABC$,$\frac{r_2(r_1+r_3)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}} =$

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