[b]Q2.[/b] Let be given a parallegogram $ABCD$ with the area of $12 \ \text{cm}^2$. The line through $A$ and the midpoint $M$ of $BC$ mects $BD$ at $N.$ Compute the area of the quadrilateral $MNDC.$
\[(A) \; 4 \text{cm}^2; \qquad (B) \; 5 \text{cm}^2; \qquad (C ) \; 6 \text{cm}^2; \qquad (D) \; 7 \text{cm}^2;  \qquad (E) \; \text{None of the above.}\]
1. Given a parallelogram \(ABCD\) with an area of \(12 \ \text{cm}^2\). The line through \(A\) and the midpoint \(M\) of \(BC\) intersects \(BD\) at \(N\). We need to find the area of the quadrilateral \(MNDC\).

2. Since \(M\) is the midpoint of \(BC\), the line \(AM\) divides the parallelogram into two equal areas. Therefore, the area of \(\triangle ABM\) is half of the area of \(\triangle ABC\), which is \(\frac{1}{2} \times 6 = 3 \ \text{cm}^2\).

3. The area of \(\triangle ABD\) is half of the area of the parallelogram \(ABCD\), so \([ABD] = \frac{1}{2} \times 12 = 6 \ \text{cm}^2\).

4. Let the area of quadrilateral \(MNDC\) be \(T\). Then the area of \(\triangle AND\) is \(9 - T\) and the area of \(\triangle MNB\) is \(6 - T\).

5. The ratio of the areas of \(\triangle AND\) to \(\triangle MNB\) is given as \(4:1\). Therefore, we have:
   \[
   \frac{9 - T}{6 - T} = 4
   \]

6. Solving the equation:
   \[
   \frac{9 - T}{6 - T} = 4 \implies 9 - T = 4(6 - T) \implies 9 - T = 24 - 4T \implies 3T = 15 \implies T = 5
   \]

7. Therefore, the area of the quadrilateral \(MNDC\) is \(5 \ \text{cm}^2\).

The final answer is \(\boxed{5 \ \text{cm}^2}\).