If a, b and c are the sides of triangle ABC f | vedclass

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(B) Given that the exradii are $r_1 = 8, r_2 = 12, r_3 = 24$. We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$,where $r$ is

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$(12, 16, 20)$

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(B) Given that the exradii are $r_1 = 8, r_2 = 12, r_3 = 24$. We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$,where $r$ is the inradius. $\frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{3+2+1}{24} = \frac{6}{24} = \frac{1}{4}$,so $r = 4$. Also,$\Delta = \sqrt{r r_1 r_2 r_3} = \sqrt{4 	imes 8 	imes 12 	imes 24} = \sqrt{9216} = 96$. Using $r_1 = \frac{\Delta}{s-a}$,we have $8 = \frac{96}{s-a} \Rightarrow s-a = 12$. Using $r_2 = \frac{\Delta}{s-b}$,we have $12 = \frac{96}{s-b} \Rightarrow s-b = 8$. Using $r_3 = \frac{\Delta}{s-c}$,we have $24 = \frac{96}{s-c} \Rightarrow s-c = 4$. Adding these: $3s - (a+b+c) = 24$ $\Rightarrow 3s - 2s = 24$ $\Rightarrow s = 24$. Then $a = s-12 = 12$,$b = s-8 = 16$,and $c = s-4 = 20$. Thus,the ordered triad $(a, b, c) = (12, 16, 20)$.

If $a, b$ and $c$ are the sides of $\triangle ABC$ for which $r_1=8, r_2=12$ and $r_3=24$,then the ordered triad $(a, b, c)=$

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