In triangle ABC,if a=8, b=10, c=12,then frac{ | vedclass

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(B) Given sides are $a=8, b=10, c=12$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{8+10+12}{2} = 15$. Area of triangle $\Delta = \sqrt{s(s-a)(s-b)(s-c

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$\frac{7}{16}$

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(B) Given sides are $a=8, b=10, c=12$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{8+10+12}{2} = 15$. Area of triangle $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{15(15-8)(15-10)(15-12)} = \sqrt{15 	imes 7 	imes 5 	imes 3} = \sqrt{1575} = 15\sqrt{7}$. Inradius $r = \frac{\Delta}{s} = \frac{15\sqrt{7}}{15} = \sqrt{7}$. Circumradius $R = \frac{abc}{4\Delta} = \frac{8 	imes 10 	imes 12}{4 	imes 15\sqrt{7}} = \frac{960}{60\sqrt{7}} = \frac{16}{\sqrt{7}}$. Therefore,$\frac{r}{R} = \frac{\sqrt{7}}{16/\sqrt{7}} = \frac{7}{16}$.

In $\triangle ABC$,if $a=8, b=10, c=12$,then $\frac{r}{R}=$

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