In $\Delta ABC$, $D$, $E$ and $F$ are respectively the mid-points of sides $BC$, $CA$ and $AB$. If the area of $\Delta ABC=16\text{sq cm}$, then the area of the trapezium $FBCE$ is:

Connecting the midpoints $D$, $E$, and $F$ splits $\Delta ABC$ into four congruent smaller triangles of equal area: $\Delta AFE$, $\Delta FBD$, $\Delta EDC$, and $\Delta DEF$.
$$\text{Area of each small triangle}=\frac{\text{Area}(\Delta ABC)}{4}=\frac{16}{4}=4\text{sq cm}$$

Trapezium $FBCE$ contains exactly three of these constituent smaller triangles ($\Delta FBD$, $\Delta DEF$, and $\Delta EDC$). Alternatively, its area can be found by subtracting the top triangle from the whole:

$$\text{Area}(FBCE)=\text{Area}(\Delta ABC)-\text{Area}(\Delta AFE)$$

$$\text{Area}(FBCE)=16-4=12\text{sq cm}$$