If the sides of a triangle are in the ratio 3 | vedclass

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(B) Let the sides of the triangle be $a = 3k, b = 7k, c = 8k$. The semi-perimeter $s$ is given by $s = \frac{a + b + c}{2} = \frac{3k + 7k + 8k}{2} =

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$7:2$

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(B) Let the sides of the triangle be $a = 3k, b = 7k, c = 8k$. The semi-perimeter $s$ is given by $s = \frac{a + b + c}{2} = \frac{3k + 7k + 8k}{2} = 9k$. The circumradius $R$ is given by $R = \frac{abc}{4\Delta}$ and the inradius $r$ is given by $r = \frac{\Delta}{s}$,where $\Delta$ is the area of the triangle. Thus,$\frac{R}{r} = \frac{abc}{4\Delta} 	imes \frac{s}{\Delta} = \frac{abc \cdot s}{4\Delta^2}$. Using Heron's formula,$\Delta^2 = s(s-a)(s-b)(s-c) = 9k(9k-3k)(9k-7k)(9k-8k) = 9k(6k)(2k)(k) = 108k^4$. Substituting these values: $\frac{R}{r} = \frac{(3k)(7k)(8k)(9k)}{4(108k^4)} = \frac{1512k^4}{432k^4} = \frac{7}{2}$. Therefore,$R : r = 7 : 2$.

If the sides of a triangle are in the ratio $3 : 7 : 8,$ then $R : r$ is equal to

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