If r is the inradius,Delta is the area of tri | vedclass

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(D) Let the lengths of the sides of $	riangle ABC$ be $a, b,$ and $c$. The semi-perimeter $s$ is given by $s = \frac{a+b+c}{2}$.From the diagram,the

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$\Delta = rs$

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(D) Let the lengths of the sides of $	riangle ABC$ be $a, b,$ and $c$. The semi-perimeter $s$ is given by $s = \frac{a+b+c}{2}$.From the diagram,the area of $	riangle ABC$ can be expressed as the sum of the areas of $	riangle IBC, 	riangle ICA,$ and $	riangle IAB$,where $I$ is the incenter:$\Delta = 	ext{Area}(	riangle IBC) + 	ext{Area}(	riangle ICA) + 	ext{Area}(	riangle IAB)$$\Delta = \frac{1}{2}ar + \frac{1}{2}br + \frac{1}{2}cr$$\Delta = r \left( \frac{a+b+c}{2} ight)$Since $s = \frac{a+b+c}{2}$,we have $\Delta = rs$.Thus,option $(d)$ is correct.

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

If $r$ is the inradius,$\Delta$ is the area of $\triangle ABC$,and $s$ is the semi-perimeter,then which of the following is true?

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