With usual notation in a Delta ABC,left( frac | vedclass

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(C) We know that $\frac{1}{r_1} + \frac{1}{r_2} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} = \frac{2s - a - b}{\Delta} = \frac{c}{\Delta}$.Similarly,$\

Vedclass · Accepted Answer

$64$

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(C) We know that $\frac{1}{r_1} + \frac{1}{r_2} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} = \frac{2s - a - b}{\Delta} = \frac{c}{\Delta}$.Similarly,$\frac{1}{r_2} + \frac{1}{r_3} = \frac{a}{\Delta}$ and $\frac{1}{r_3} + \frac{1}{r_1} = \frac{b}{\Delta}$.Therefore,the $LHS = \left( \frac{c}{\Delta} \right) \left( \frac{a}{\Delta} \right) \left( \frac{b}{\Delta} \right) = \frac{abc}{\Delta^3}$.Using the relation $\Delta = \frac{abc}{4R}$,we have $\Delta^3 = \frac{a^3 b^3 c^3}{64 R^3}$.Substituting this into the $LHS$,we get $\frac{abc}{\frac{a^3 b^3 c^3}{64 R^3}} = \frac{64 R^3}{a^2 b^2 c^2}$.Comparing this with $\frac{K R^3}{a^2 b^2 c^2}$,we find $K = 64$.

With usual notation in a $\Delta ABC$,$\left( \frac{1}{r_1} + \frac{1}{r_2} \right) \left( \frac{1}{r_2} + \frac{1}{r_3} \right) \left( \frac{1}{r_3} + \frac{1}{r_1} \right) = \frac{K R^3}{a^2 b^2 c^2}$ where $K$ has the value equal to

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