In triangle ABC, if a : b : c = 4 : 5 : 6, th | vedclass

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(A) Given $a : b : c = 4 : 5 : 6$. Let $a = 4k, b = 5k, c = 6k$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{15k}{2}$. Area $\Delta = \sqrt{s(s-a)(s-b

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$16 : 7$

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(A) Given $a : b : c = 4 : 5 : 6$. Let $a = 4k, b = 5k, c = 6k$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{15k}{2}$. Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15k}{2}(\frac{7k}{2})(\frac{5k}{2})(\frac{3k}{2})} = \frac{15\sqrt{7}k^2}{4}$. Circumradius $R = \frac{abc}{4\Delta} = \frac{120k^3}{4 	imes \frac{15\sqrt{7}k^2}{4}} = \frac{8k}{\sqrt{7}}$. Inradius $r = \frac{\Delta}{s} = \frac{15\sqrt{7}k^2}{4} 	imes \frac{2}{15k} = \frac{\sqrt{7}k}{2}$. Ratio $\frac{R}{r} = \frac{8k}{\sqrt{7}} 	imes \frac{2}{\sqrt{7}k} = \frac{16}{7}$.

In $\triangle ABC$, if $a : b : c = 4 : 5 : 6$, then the ratio of the circumradius to its inradius is

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