In triangle ABC,if r_1=3, r_2=10 and r_3=15,t | vedclass

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(C) We know the relation between the exradii $r_1, r_2, r_3$ and the circumradius $R$ is given by $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \fr

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$\frac{13}{2}$

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(C) We know the relation between the exradii $r_1, r_2, r_3$ and the circumradius $R$ is given by $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$,where $r$ is the inradius. Also,the area of the triangle $\Delta = \sqrt{r r_1 r_2 r_3}$. Given $r_1=3, r_2=10, r_3=15$. $\frac{1}{r} = \frac{1}{3} + \frac{1}{10} + \frac{1}{15} = \frac{10+3+2}{30} = \frac{15}{30} = \frac{1}{2}$. So,$r=2$. Now,$\Delta = \sqrt{2 	imes 3 	imes 10 	imes 15} = \sqrt{900} = 30$. We also have the formula $\Delta = rs$,where $s$ is the semi-perimeter. $30 = 2s \implies s = 15$. Using the formula $r_1 = \frac{\Delta}{s-a}$,we get $3 = \frac{30}{15-a} \implies 15-a = 10 \implies a = 5$. Similarly,$r_2 = \frac{\Delta}{s-b} \implies 10 = \frac{30}{15-b} \implies 15-b = 3 \implies b = 12$. And $r_3 = \frac{\Delta}{s-c} \implies 15 = \frac{30}{15-c} \implies 15-c = 2 \implies c = 13$. Since $a^2 + b^2 = 5^2 + 12^2 = 25 + 144 = 169 = 13^2 = c^2$,the triangle is a right-angled triangle. For a right-angled triangle,the circumradius $R = \frac{c}{2} = \frac{13}{2} = 6.5$.

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 $\triangle ABC$,if $r_1=3, r_2=10$ and $r_3=15$,then $R=$

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