In a triangle,if the ex-radii r_1, r_2, r_3 a | vedclass

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(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\Delta$ is the area of the triangle and $

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$5: 8: 9$

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(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\Delta$ is the area of the triangle and $s$ is the semi-perimeter.Given $r_1: r_2: r_3 = 1: 2: 3$,let $r_1 = x, r_2 = 2x, r_3 = 3x$.Then $s-a = \frac{\Delta}{x}$,$s-b = \frac{\Delta}{2x}$,and $s-c = \frac{\Delta}{3x}$.Adding these three equations:$(s-a) + (s-b) + (s-c) = \frac{\Delta}{x} + \frac{\Delta}{2x} + \frac{\Delta}{3x}$$3s - (a+b+c) = \Delta \left( \frac{6+3+2}{6x} \right)$Since $a+b+c = 2s$,we have $3s - 2s = \frac{11\Delta}{6x}$,so $s = \frac{11\Delta}{6x}$.Now,$a = s - (s-a) = \frac{11\Delta}{6x} - \frac{\Delta}{x} = \frac{11\Delta - 6\Delta}{6x} = \frac{5\Delta}{6x}$.$b = s - (s-b) = \frac{11\Delta}{6x} - \frac{\Delta}{2x} = \frac{11\Delta - 3\Delta}{6x} = \frac{8\Delta}{6x}$.$c = s - (s-c) = \frac{11\Delta}{6x} - \frac{\Delta}{3x} = \frac{11\Delta - 2\Delta}{6x} = \frac{9\Delta}{6x}$.Thus,$a: b: c = \frac{5\Delta}{6x} : \frac{8\Delta}{6x} : \frac{9\Delta}{6x} = 5: 8: 9$.

In a triangle,if the ex-radii $r_1, r_2, r_3$ are in the ratio $1: 2: 3$,then its sides are in the ratio

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