$D, E$ and $F$ are respectively the mid-points of sides $AB, BC$ and $CA$ of $\Delta ABC$. Find the ratio of the areas of $\Delta DEF$ and $\Delta ABC$.

(1:4) Given that $D, E$ and $F$ are the mid-points of sides $AB, BC$ and $CA$ respectively in $\Delta ABC$.
By the Mid-point Theorem,the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.
Therefore,$DE \parallel AC$ and $DE = \frac{1}{2} AC$.
Similarly,$EF \parallel AB$ and $EF = \frac{1}{2} AB$,and $DF \parallel BC$ and $DF = \frac{1}{2} BC$.
In $\Delta DEF$ and $\Delta ABC$:
$\frac{DE}{AC} = \frac{EF}{AB} = \frac{DF}{BC} = \frac{1}{2}$.
By $SSS$ similarity criterion,$\Delta DEF \sim \Delta ABC$.
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\operatorname{ar}(\Delta DEF)}{\operatorname{ar}(\Delta ABC)} = \left(\frac{DE}{AC}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Thus,the ratio of the areas of $\Delta DEF$ and $\Delta ABC$ is $1:4$.