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

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(B) Given $r_1=4, r_2=8, r_3=24$. We know that $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{4} + \frac{1}{8} + \frac{1}{24}

Vedclass · Accepted Answer

$\frac{16}{\sqrt{5}}$

Vedclass · Answer

(B) Given $r_1=4, r_2=8, r_3=24$. We know that $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{4} + \frac{1}{8} + \frac{1}{24} = \frac{6+3+1}{24} = \frac{10}{24} = \frac{5}{12}$. Thus,$r = \frac{12}{5}$. We know that $\Delta^2 = r r_1 r_2 r_3 = \frac{12}{5} 	imes 4 	imes 8 	imes 24 = \frac{9216}{5}$. So,$\Delta = \sqrt{\frac{9216}{5}} = \frac{96}{\sqrt{5}}$. Using $r = \frac{\Delta}{s}$,we get $s = \frac{\Delta}{r} = \frac{96}{\sqrt{5}} 	imes \frac{5}{12} = 8\sqrt{5}$. Since $r_1 = \frac{\Delta}{s-a}$,we have $4 = \frac{96/\sqrt{5}}{8\sqrt{5}-a}$. $4(8\sqrt{5}-a) = \frac{96}{\sqrt{5}} \Rightarrow 32\sqrt{5} - 4a = \frac{96}{\sqrt{5}}$. $4a = 32\sqrt{5} - \frac{96}{\sqrt{5}} = \frac{160-96}{\sqrt{5}} = \frac{64}{\sqrt{5}}$. $a = \frac{16}{\sqrt{5}}$.

In $\triangle ABC$,if $r_1=4, r_2=8, r_3=24$,then $a=$

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