Mathematics - Properties of Triangles Question with Solution | TestHub

We have $\Delta =\frac{\sqrt{3}}{4}{\mathrm{a}}^2,\mathrm{s}=\frac{3\mathrm{a}}{2}$ $\therefore r=\frac{\Delta }{\mathrm{s}}=\frac{\mathrm{a}}{2\sqrt{3}},\mathrm{R}=\frac{\mathrm{abc}}{4\Delta }=\frac{{\mathrm{a}}^3}{\sqrt{3}{\mathrm{a}}^2}=\frac{\mathrm{a}}{\sqrt{3}}$ and $r_1=\frac{\Delta }{s-a}=\frac{\sqrt{3}/4a^2}{a/2}=\frac{\sqrt{3}}{2}a$ Hence, $\mathrm{r}:\mathrm{R}:{\mathrm{r}}_1=\frac{\mathrm{a}}{2\sqrt{3}}:\frac{\mathrm{a}}{\sqrt{3}}:\frac{\sqrt{3}}{2}\mathrm{a}=1:2:3$