c
$\operatorname{Let} \mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right), \mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ and $\mathrm{C}\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right)$ be the vertices
of a $\Delta \mathrm{ABC},$ and let $\mathrm{G}$ be its centroid. Then, $\Delta_{1}=$ Area of $\Delta \mathrm{GBC}$
$\Rightarrow \Delta_{1}=\frac{\Delta}{3},$ where $\Delta$ is the area of $\Delta \mathrm{ABC}$
$\Delta_{2}=$ Area of triangle formed by the mid-points of the sides
$\Rightarrow \Delta_{2}=\frac{1}{4} \Delta$
$\therefore \Delta_{1}: \Delta_{2}=4: 3$