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

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(A) Given $a=4k, b=5k, c=6k$. 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} 	ime

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$16:7$

Vedclass · Answer

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

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

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