In triangle ABC,if a=6, b=8 and c=10,then fra | vedclass

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(A) Given $a=6, b=8, c=10$. Since $6^2 + 8^2 = 36 + 64 = 100 = 10^2$,the triangle is a right-angled triangle with the hypotenuse $c=10$.The semi-perim

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$b+c$

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(A) Given $a=6, b=8, c=10$. Since $6^2 + 8^2 = 36 + 64 = 100 = 10^2$,the triangle is a right-angled triangle with the hypotenuse $c=10$.The semi-perimeter $s = \frac{a+b+c}{2} = \frac{6+8+10}{2} = 12$.The area of the triangle $\Delta = \frac{1}{2} 	imes 6 	imes 8 = 24$.The exradii are given by $r_1 = \frac{\Delta}{s-a} = \frac{24}{12-6} = 4$,$r_2 = \frac{\Delta}{s-b} = \frac{24}{12-8} = 6$,$r_3 = \frac{\Delta}{s-c} = \frac{24}{12-10} = 12$.The inradius $r = \frac{\Delta}{s} = \frac{24}{12} = 2$.Now,calculate the expression $\frac{2 r_2 r_3}{r r_1} = \frac{2 	imes 6 	imes 12}{2 	imes 4} = \frac{144}{8} = 18$.Check the options: $a+b = 6+8 = 14$,$b+c = 8+10 = 18$,$c+a = 10+6 = 16$,$a+b+c = 24$.Thus,the value is $b+c = 18$.

In triangle $ABC$,if $a=6, b=8$ and $c=10$,then $\frac{2 r_2 r_3}{r r_1} = $

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