In a triangle ABC,if r_1=4, r_2=8 and r_3=24, | vedclass

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(A) We know that the exradii of a triangle are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Given $r_

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$4: 7: 9$

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(A) We know that the exradii of a triangle are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Given $r_1=4, r_2=8, r_3=24$.Taking reciprocals,we have $\frac{1}{r_1} = \frac{s-a}{\Delta} = \frac{1}{4}$,$\frac{1}{r_2} = \frac{s-b}{\Delta} = \frac{1}{8}$,and $\frac{1}{r_3} = \frac{s-c}{\Delta} = \frac{1}{24}$.Adding these,$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{3s-(a+b+c)}{\Delta} = \frac{3s-2s}{\Delta} = \frac{s}{\Delta} = \frac{1}{r}$.So,$\frac{1}{r} = \frac{1}{4} + \frac{1}{8} + \frac{1}{24} = \frac{6+3+1}{24} = \frac{10}{24} = \frac{5}{12}$,which means $r = \frac{12}{5}$.Now,$\frac{s-a}{\Delta} = \frac{1}{4} \implies s-a = \frac{\Delta}{4} = \frac{rs}{4} = \frac{(12/5)s}{4} = \frac{3s}{5} \implies a = s - \frac{3s}{5} = \frac{2s}{5}$.Similarly,$s-b = \frac{\Delta}{8} = \frac{(12/5)s}{8} = \frac{3s}{10} \implies b = s - \frac{3s}{10} = \frac{7s}{10}$.And $s-c = \frac{\Delta}{24} = \frac{(12/5)s}{24} = \frac{s}{10} \implies c = s - \frac{s}{10} = \frac{9s}{10}$.Thus,$a:b:c = \frac{2s}{5} : \frac{7s}{10} : \frac{9s}{10} = 4:7:9$.

In a triangle $ABC$,if $r_1=4, r_2=8$ and $r_3=24$,then $a: b: c=$

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