In a triangle ABC, a:b:c = 4:5:6. The ratio o | vedclass

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

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$\frac{16}{7}$

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

In a triangle $ABC,$ $a:b:c = 4:5:6$. The ratio of the radius of the circumcircle to that of the incircle is

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