If $D,E,F$ are respectively the mid points of the sides $BC,CA$ and $AB$ of $△ABC$ and the area of $△ABC$ is $24$ sq. cm, then the area of $△DEF$ is -

The correct option is C: $6c{m}^2$

Given: $D,E$ and $F$ are the mid-points of sides $BC,AC$ and $AB$ respectively of $\Delta ABC$.

Mid Point Theorem -The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.

$FE=\frac{1}{2}BC$ [Using above theorem]

$\frac{FE}{BC}=\frac{1}{2}$

Similarly. $DE=\frac{1}{2}AB$ [Using above theorem]

$\frac{DE}{AB}=\frac{1}{2}$

and, $\frac{FD}{AC}=\frac{1}{2}$

$\therefore \frac{FE}{BC}=\frac{DE}{AB}=\frac{FD}{AC}$

Thus, $\Delta ABC\sim \Delta DEF$ [by SSS similarity rule]

The ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

$\frac{Area\Delta DEF}{Area\Delta ABC}={\left(\frac{DE}{AB}\right)}^2$

$\frac{Area\Delta DEF}{Area\Delta ABC}={\left(\frac{DE}{2DE}\right)}^2$

Clearly, $area\left(△DEF\right)=\frac{1}{4}\times area\left(△ABC\right)$

$=(\frac{1}{4}\times 24)=6c{m}^2$