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

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(D) Given,$\frac{a}{4} = \frac{b}{5} = \frac{c}{6} = k$ (let). So,$a = 4k, b = 5k, c = 6k$. The semi-perimeter $s = \frac{a+b+c}{2} = \frac{15k}{2}$.

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$16: 7$

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(D) Given,$\frac{a}{4} = \frac{b}{5} = \frac{c}{6} = k$ (let). So,$a = 4k, b = 5k, c = 6k$. The semi-perimeter $s = \frac{a+b+c}{2} = \frac{15k}{2}$. The area of the triangle $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15k}{2} \cdot \frac{7k}{2} \cdot \frac{5k}{2} \cdot \frac{3k}{2}} = \frac{15\sqrt{7}k^2}{4}$. The inradius $r = \frac{\Delta}{s} = \frac{15\sqrt{7}k^2 / 4}{15k / 2} = \frac{\sqrt{7}k}{2}$. The circumradius $R = \frac{abc}{4\Delta} = \frac{(4k)(5k)(6k)}{4 \cdot (15\sqrt{7}k^2 / 4)} = \frac{120k^3}{15\sqrt{7}k^2} = \frac{8k}{\sqrt{7}}$. Therefore,the ratio $\frac{R}{r} = \frac{8k}{\sqrt{7}} \div \frac{\sqrt{7}k}{2} = \frac{8k}{\sqrt{7}} \cdot \frac{2}{\sqrt{7}k} = \frac{16}{7}$.

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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