In a triangle ABC , if a:b:c =4:5:6, then the ratio of the radius of the circumcircle to that of the': incircle is

$\begin{array}{l}\text{ Thet }\mathrm{a}=4\mathrm{\lambda },\mathrm{b}=5\mathrm{\lambda },\mathrm{c}=6\mathrm{\lambda };\mathrm{R}=\frac{\mathrm{abc}}{4\mathrm{\Delta }},\mathrm{r}=\frac{\mathrm{\Delta }}{\mathrm{s}}\\ \mathrm{then}\frac{\mathrm{R}}{\mathrm{r}}=\frac{\mathrm{abc}}{4\mathrm{\Delta }}\times \frac{\mathrm{s}}{\mathrm{\Delta }}=\frac{\mathrm{sabc}}{4{\mathrm{\Delta }}^2}\\ =\frac{\mathrm{sabc}}{4\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\end{array}$
$\begin{array}{l}=\frac{4\lambda \times 5\lambda \times 6\lambda }{4\left(\frac{15}{2}\lambda -4\lambda \right)\left(\frac{15}{2}\lambda -5\lambda \right)\left(\frac{15}{2}\lambda -6\lambda \right)}\\ =\frac{5\times 6\times 8}{7\times 5\times 3}=\frac{16}{7}\therefore \mathrm{R}:r=16:7\end{array}$