In a triangle ABC, r_1, r_2 and r_3 respectiv | vedclass

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(B) We know that the area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$. Thus, $\frac{1}{p_1} = \frac{a}{2\Delt

Vedclass · Accepted Answer

$\frac{1}{r}\left(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\right)$

Vedclass · Answer

(B) We know that the area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$. Thus, $\frac{1}{p_1} = \frac{a}{2\Delta}$, $\frac{1}{p_2} = \frac{b}{2\Delta}$, and $\frac{1}{p_3} = \frac{c}{2\Delta}$. Therefore, $\left(\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}\right) = \frac{a+b+c}{2\Delta} = \frac{2s}{2\Delta} = \frac{s}{\Delta} = \frac{1}{r}$. Squaring both sides, we get $\left(\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}\right)^2 = \frac{1}{r^2}$. We also know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$. Thus, $\frac{1}{r^2} = \frac{1}{r} \left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\right)$. Hence, the correct option is $B$.

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