In a parallelogram ABCD, point P lies in DC such that DP:PC= `3:2`. If area of `DeltaDPB=30sq.` cm, ffind the area of the parallelogram ABCD.

Answer

To find the area of the parallelogram ABCD given that point P lies on DC such that DP:PC = 3:2 and the area of triangle DPB is 30 cm², we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Ratios**:
- We know that DP:PC = 3:2. This means if we let DP = 3x and PC = 2x for some value x, then the total length of DC = DP + PC = 3x + 2x = 5x.

2. **Area of Triangle DPB**:
- The area of triangle DPB is given as 30 cm². We can denote the height from point B to line DC as h1.

3. **Finding Area of Triangle BPC**:
- Since the triangles DPB and BPC share the same height (h1) from point B to line DC, we can express the area of triangle BPC in terms of its base PC.
- The area of triangle BPC can be calculated using the ratio of the bases:
\[ \frac{\text{Area of } \triangle DPB}{\text{Area of } \triangle BPC} = \frac{DP}{PC} \]
- Substituting the known values:
\[ \frac{30}{\text{Area of } \triangle BPC} = \frac{3}{2} \]
- Cross-multiplying gives:
\[ 3 \cdot \text{Area of } \triangle BPC = 30 \cdot 2 \]
\[ 3 \cdot \text{Area of } \triangle BPC = 60 \]
\[ \text{Area of } \triangle BPC = \frac{60}{3} = 20 \text{ cm}^2 \]

4. **Finding Area of Triangle BDC**:
- The area of triangle BDC can be found by adding the areas of triangles DPB and BPC:
\[ \text{Area of } \triangle BDC = \text{Area of } \triangle DPB + \text{Area of } \triangle BPC \]
\[ \text{Area of } \triangle BDC = 30 \text{ cm}^2 + 20 \text{ cm}^2 = 50 \text{ cm}^2 \]

5. **Finding Area of Parallelogram ABCD**:
- The area of parallelogram ABCD is twice the area of triangle BDC (since diagonal BD divides the parallelogram into two equal areas):
\[ \text{Area of parallelogram ABCD} = 2 \times \text{Area of } \triangle BDC \]
\[ \text{Area of parallelogram ABCD} = 2 \times 50 \text{ cm}^2 = 100 \text{ cm}^2 \]

### Final Answer:
The area of parallelogram ABCD is **100 cm²**.