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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Aicharya · Accepted Answer

Given that, $r_1, r_2, r_3$ are ex-radii of triangle and $ \begin{aligned} & r_1: r_2: r_3=1: 2: 3 \\ & r_1=x, r_2=2 x, r_3=3 x \end{aligned} $

On adding Eqs. (i), (ii) and (iii), we get $ \begin{gathered} 3 s-(a+b+c)=\frac{\Delta}{x}+\frac{\Delta}{2 x}+\frac{\Delta}{3 x} \\ s=\frac{11 \Delta}{6 x} \end{gathered} $ From Eqs. (i), (ii), (iii), we get $ a=\frac{5 \Delta}{6 x}, b=\frac{8 \Delta}{6 x}, c=\frac{9 \Delta}{6 x} $ So, $\quad a: b: c=5: 8: 9$

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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