In a triangle ABC,let a, b, c, s, r, R, I, S, | vedclass

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(C) . $\because 	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \frac{(s-b)(s-c)}{\Delta}$.Also,$\frac{r}{s-a} = \frac{\Delta}{s(s-a)} = \sqrt{\

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(C) . $\because 	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \frac{(s-b)(s-c)}{\Delta}$.Also,$\frac{r}{s-a} = \frac{\Delta}{s(s-a)} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \frac{(s-b)(s-c)}{\Delta}$.$	herefore 	an \frac{A}{2} = \frac{r}{s-a} = \frac{(s-b)(s-c)}{\Delta}$. Thus,$A-V$.$B$. From the figure,$r = IF = (AI) \sin \frac{A}{2} = (AI) \sqrt{\frac{(s-b)(s-c)}{bc}}$. Thus,$B-I$.$C$. $SI$ is the distance between the circumcentre and in-centre,given by $SI = \sqrt{R^2 - 2rR}$.Squaring gives $(SI)^2 = R^2 - 2rR$,so $(SI)^2 + 2Rr = R^2$. Thus,$C-II$.$D$. We know $r_1 + r_2 + r_3 = 4R + r$.Squaring both sides: $r_1^2 + r_2^2 + r_3^2 = (4R + r)^2 - 2(r_1r_2 + r_2r_3 + r_3r_1)$.Using $r_1r_2 + r_2r_3 + r_3r_1 = s^2$,we get $r_1^2 + r_2^2 + r_3^2 = (4R + r)^2 - 2s^2 = (4R + r + \sqrt{2}s)(4R + r - \sqrt{2}s)$. Thus,$D-III$.

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

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