ABCD is a parallelogram. P and Q are mid points of BC & CD. Find the area of `DeltaAPQ` if area of `DeltaABC` is 12.

Answer

To find the area of triangle \( \Delta APQ \) in the parallelogram \( ABCD \) where \( P \) and \( Q \) are the midpoints of sides \( BC \) and \( CD \) respectively, and the area of triangle \( \Delta ABC \) is given as 12, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Parallelogram**:
- We know that in a parallelogram, the area of triangles formed by its diagonals and midpoints can be related to the area of the entire parallelogram.

2. **Area of Parallelogram**:
- The area of parallelogram \( ABCD \) is twice the area of triangle \( ABC \).
- Given that the area of triangle \( ABC \) is 12, the area of parallelogram \( ABCD \) is:
\[ \text{Area of } ABCD = 2 \times \text{Area of } ABC = 2 \times 12 = 24 \]

3. **Finding Areas of Smaller Triangles**:
- Since \( P \) and \( Q \) are midpoints, triangles \( ADQ \) and \( ABP \) each have half the area of triangle \( ABC \).
- Therefore, the area of triangle \( ADQ \) is:
\[ \text{Area of } ADQ = \frac{1}{2} \times \text{Area of } ABC = \frac{1}{2} \times 12 = 6 \]
- Similarly, the area of triangle \( ABP \) is also:
\[ \text{Area of } ABP = 6 \]

4. **Finding Area of Triangle \( BDC \)**:
- The area of triangle \( BDC \) is equal to the area of the parallelogram minus the area of triangles \( ADQ \) and \( ABP \):
\[ \text{Area of } BDC = \text{Area of } ABCD - (\text{Area of } ADQ + \text{Area of } ABP) \]
\[ \text{Area of } BDC = 24 - (6 + 6) = 24 - 12 = 12 \]

5. **Finding Area of Triangle \( QCP \)**:
- Since \( P \) and \( Q \) are midpoints, triangle \( QCP \) will have an area that is half of triangle \( BDC \):
\[ \text{Area of } QCP = \frac{1}{2} \times \text{Area of } BDC = \frac{1}{2} \times 12 = 6 \]

6. **Finding Area of Triangle \( APQ \)**:
- The area of triangle \( APQ \) can now be calculated by subtracting the areas of triangles \( ADQ \), \( ABP \), and \( QCP \) from the total area of the parallelogram:
\[ \text{Area of } APQ = \text{Area of } ABCD - (\text{Area of } ADQ + \text{Area of } ABP + \text{Area of } QCP) \]
\[ \text{Area of } APQ = 24 - (6 + 6 + 6) = 24 - 18 = 6 \]

### Final Result:
The area of triangle \( \Delta APQ \) is \( 6 \, \text{cm}^2 \).