Given: In $△ABC,$  D,E and F are midpoints of BC, CA and AB.
Area (ΔABC) = 32 cm²
To find: Area of trapezium BFEC

Consider $△ABC,$
F and E are midpoints of AB and AC. (given)
$\therefore$  FE $\parallel BC$ (Midpoint theorem)
$\therefore$  FE $\parallel BD$
Similarly ED $\parallel$ AB and FD $\parallel$ AC
$\therefore$ FEDB, FDEC and FDEA are all parallelograms.
Since a diagonal divides a parallelogram into two congruent triangles, hence
$Area\left(\Delta BFD\right)=Area\left(\Delta EFD\right)=Area\left(\Delta ECD\right)=Area\left(\Delta EFA\right)$
$=\frac{1}{4}Area\left(\Delta ABC\right)=8c{m}^2$
Area(BFEC)
$=Area\left(\Delta BFD\right)+Area\left(\Delta EFD\right)+Area\left(\Delta ECD\right)=24c{m}^2$