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

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

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$\frac{16}{7}$

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(A) Given $a: b: c = 4: 5: 6$. Let $a = 4k, b = 5k, c = 6k$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{4k+5k+6k}{2} = \frac{15k}{2}$. Using Heron's formula,the area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15k}{2} \cdot \frac{7k}{2} \cdot \frac{5k}{2} \cdot \frac{3k}{2}} = \sqrt{\frac{15 \cdot 7 \cdot 5 \cdot 3}{16} k^4} = \sqrt{\frac{1575}{16} k^4} = \frac{15\sqrt{7}}{4} k^2$. Circumradius $R = \frac{abc}{4\Delta} = \frac{(4k)(5k)(6k)}{4 \cdot \frac{15\sqrt{7}}{4} k^2} = \frac{120k^3}{15\sqrt{7}k^2} = \frac{8}{\sqrt{7}} k$. Inradius $r = \frac{\Delta}{s} = \frac{\frac{15\sqrt{7}}{4} k^2}{\frac{15k}{2}} = \frac{15\sqrt{7}}{4} \cdot \frac{2}{15} k = \frac{\sqrt{7}}{2} k$. Ratio $R: r = \frac{8}{\sqrt{7}} k : \frac{\sqrt{7}}{2} k = \frac{8}{\sqrt{7}} : \frac{\sqrt{7}}{2} = \frac{16}{7}$.

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

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