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

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(A) Let $a = 4K, b = 5K, c = 6K$.The semi-perimeter $s = \frac{a+b+c}{2} = \frac{15K}{2}$.The circumradius $R = \frac{abc}{4\Delta}$ and the inradius

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$\frac{16}{7}$

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(A) Let $a = 4K, b = 5K, c = 6K$.The semi-perimeter $s = \frac{a+b+c}{2} = \frac{15K}{2}$.The circumradius $R = \frac{abc}{4\Delta}$ and the inradius $r = \frac{\Delta}{s}$.Therefore,the ratio $\frac{R}{r} = \frac{abc}{4\Delta} 	imes \frac{s}{\Delta} = \frac{abcs}{4\Delta^2}$.Using Heron's formula,$\Delta^2 = s(s-a)(s-b)(s-c)$.$\frac{R}{r} = \frac{abcs}{4s(s-a)(s-b)(s-c)} = \frac{abc}{4(s-a)(s-b)(s-c)}$.Substituting the values:$s-a = \frac{15K}{2} - 4K = \frac{7K}{2}$$s-b = \frac{15K}{2} - 5K = \frac{5K}{2}$$s-c = \frac{15K}{2} - 6K = \frac{3K}{2}$$\frac{R}{r} = \frac{(4K)(5K)(6K)}{4(\frac{7K}{2})(\frac{5K}{2})(\frac{3K}{2})} = \frac{120K^3}{4(\frac{105K^3}{8})} = \frac{120K^3}{\frac{105K^3}{2}} = \frac{240}{105} = \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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