In triangle ABC,frac{1}{r_1^2}+frac{1}{r_2^2} | vedclass

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(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.Therefore,$\frac{1}{r_1}

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$\frac{a^2+b^2+c^2}{\Delta^2}$

Vedclass · Answer

(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.Therefore,$\frac{1}{r_1} = \frac{s-a}{\Delta}$,$\frac{1}{r_2} = \frac{s-b}{\Delta}$,$\frac{1}{r_3} = \frac{s-c}{\Delta}$,and $\frac{1}{r} = \frac{s}{\Delta}$.Now,$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2} = \frac{(s-a)^2}{\Delta^2} + \frac{(s-b)^2}{\Delta^2} + \frac{(s-c)^2}{\Delta^2} + \frac{s^2}{\Delta^2}$.$= \frac{(s^2+a^2-2as) + (s^2+b^2-2sb) + (s^2+c^2-2sc) + s^2}{\Delta^2}$.$= \frac{4s^2 + a^2+b^2+c^2 - 2s(a+b+c)}{\Delta^2}$.Since $a+b+c = 2s$,we have $2s(a+b+c) = 2s(2s) = 4s^2$.$= \frac{4s^2 + a^2+b^2+c^2 - 4s^2}{\Delta^2} = \frac{a^2+b^2+c^2}{\Delta^2}$.

List-$I$	List-$II$
$A. \tan \frac{A}{2} = \frac{r}{s-a}$	$I. (AI) \left( \frac{\sqrt{(s-b)(s-c)}}{bc} \right)$
$B. r$	$II. R^2$
$C. (SI)^2 + 2Rr$	$III. (4R + r + \sqrt{2}s)(4R + r - \sqrt{2}s)$
$D. r_1^2 + r_2^2 + r_3^2$	$IV. \frac{Rr}{S}$
	$V. \frac{(s-b)(s-c)}{\Delta}$

In $\triangle ABC$,$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2} =$

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