If a, b, c are the sides of a Delta ABC and e | vedclass

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(B) Given: $r_1=12, r_2=6, r_3=4$. We know that $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$. Thus, $s-a = \frac{\De

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$44$

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(B) Given: $r_1=12, r_2=6, r_3=4$. We know that $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$. Thus, $s-a = \frac{\Delta}{12}, s-b = \frac{\Delta}{6}, s-c = \frac{\Delta}{4}$. Adding these, $(s-a)+(s-b)+(s-c) = 3s-(a+b+c) = s = \Delta(\frac{1}{12} + \frac{1}{6} + \frac{1}{4}) = \Delta(\frac{1+2+3}{12}) = \frac{6\Delta}{12} = \frac{\Delta}{2}$. So, $\Delta = 2s$. Now, $s-a = \frac{2s}{12} = \frac{s}{6} \Rightarrow a = s - \frac{s}{6} = \frac{5s}{6}$. $s-b = \frac{2s}{6} = \frac{s}{3} \Rightarrow b = s - \frac{s}{3} = \frac{2s}{3}$. $s-c = \frac{2s}{4} = \frac{s}{2} \Rightarrow c = s - \frac{s}{2} = \frac{s}{2}$. Using $\Delta^2 = s(s-a)(s-b)(s-c)$, we have $(2s)^2 = s(\frac{s}{6})(\frac{s}{3})(\frac{s}{2})$. $4s^2 = \frac{s^4}{36}$ $\Rightarrow s^2 = 144$ $\Rightarrow s = 12$. Then $a = \frac{5 	imes 12}{6} = 10, b = \frac{2 	imes 12}{3} = 8, c = \frac{12}{2} = 6$. Therefore, $a+2b+3c = 10 + 2(8) + 3(6) = 10 + 16 + 18 = 44$.

If $a, b, c$ are the sides of a $\Delta ABC$ and exradii $r_1, r_2, r_3$ are respectively $12, 6, 4$, then $a+2b+3c=$

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