In triangle ABC,if AB: BC: CA = 6: 4: 5,then | vedclass

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(B) Let the sides be $c = 6k$,$a = 4k$,and $b = 5k$. The semi-perimeter $s = \frac{6k + 4k + 5k}{2} = \frac{15k}{2}$. Using Heron's formula,the area $

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$16: 7$

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(B) Let the sides be $c = 6k$,$a = 4k$,and $b = 5k$. The semi-perimeter $s = \frac{6k + 4k + 5k}{2} = \frac{15k}{2}$. Using Heron's formula,the area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15k}{2} 	imes \frac{7k}{2} 	imes \frac{5k}{2} 	imes \frac{3k}{2}} = \sqrt{\frac{1575k^4}{16}} = \frac{15\sqrt{7}k^2}{4}$. The circumradius $R = \frac{abc}{4\Delta} = \frac{(6k)(4k)(5k)}{4 	imes \frac{15\sqrt{7}k^2}{4}} = \frac{120k^3}{15\sqrt{7}k^2} = \frac{8k}{\sqrt{7}}$. The inradius $r = \frac{\Delta}{s} = \frac{15\sqrt{7}k^2}{4} 	imes \frac{2}{15k} = \frac{\sqrt{7}k}{2}$. Therefore,$\frac{R}{r} = \frac{8k}{\sqrt{7}} 	imes \frac{2}{\sqrt{7}k} = \frac{16}{7}$.

In $\triangle ABC$,if $AB: BC: CA = 6: 4: 5$,then $R: r =$

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