In triangle ABC,if r_1=8, r_2=12 and r_3=24,t | vedclass

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(B) We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$.Given $r_1=8, r_2=12, r_3=24$,we have $\frac{1}{r} = \frac{1}{8} + \fra

Vedclass · Accepted Answer

$(12, 16, 20)$

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(B) We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$.Given $r_1=8, r_2=12, r_3=24$,we have $\frac{1}{r} = \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{3+2+1}{24} = \frac{6}{24} = \frac{1}{4}$.Thus,$r=4$.Also,$r_1 = \frac{\Delta}{s-a} \implies 8 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b} \implies 12 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c} \implies 24 = \frac{\Delta}{s-c}$.Since $\Delta = rs = 4s$,we have $s-a = \frac{4s}{8} = \frac{s}{2} \implies a = \frac{s}{2}$.$s-b = \frac{4s}{12} = \frac{s}{3} \implies b = \frac{2s}{3}$.$s-c = \frac{4s}{24} = \frac{s}{6} \implies c = \frac{5s}{6}$.Using $s = \frac{a+b+c}{2} = \frac{1}{2}(\frac{s}{2} + \frac{2s}{3} + \frac{5s}{6}) = \frac{1}{2}(\frac{3s+4s+5s}{6}) = s$,which is consistent.Using $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = 4s$,we get $\sqrt{s(\frac{s}{2})(\frac{s}{3})(\frac{s}{6})} = 4s$.$\sqrt{\frac{s^4}{36}} = 4s \implies \frac{s^2}{6} = 4s \implies s = 24$.Then $a = \frac{24}{2} = 12$,$b = \frac{2(24)}{3} = 16$,$c = \frac{5(24)}{6} = 20$.So,$(a, b, c) = (12, 16, 20)$.

In $\triangle ABC$,if $r_1=8, r_2=12$ and $r_3=24$,then the ordered triple $(a, b, c) =$

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