Correct Answer: 192

Solution :
Given: $\mathrm{AD:DB=1:4}$ and the area of $\Delta \mathrm{ABC}$ is $200\;\mathrm{cm^2}$.
In $\Delta \mathrm{ABC}$,
$\mathrm{DE}\parallel\mathrm{BC}$
So, $\Delta \mathrm{ADE}\sim\Delta \mathrm{ABC}$,
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
⇒ $\frac{\text{Area of $\Delta \mathrm{ADE}$}}{{\text{Area of $\Delta \mathrm{ABC}$}}}=(\frac{\mathrm{AD}}{\mathrm{AB}})^2$
⇒ $\frac{\text{Area of $\Delta \mathrm{ADE}$}}{{200}}=(\frac{\mathrm{1}}{\mathrm{5}})^2$
⇒ $\frac{\text{Area of $\Delta \mathrm{ADE}$}}{{200}}=(\frac{\mathrm{1}}{\mathrm{25}})$
⇒ $\text{Area of $\Delta \mathrm{ADE}$}=8\;\mathrm{cm^2}$
The area of quadrilateral $\mathrm{DECB}=\text{Area of $\Delta \mathrm{ABC}$}-\text{Area of $\Delta \mathrm{ADE}$}$
$=200\;\mathrm{cm^2}-8\;\mathrm{cm^2}=192\;\mathrm{cm^2}$
Hence, the correct answer is $192$.