In `Delta ABC`, P is a point on BC such that `BP:PC=4:5` and Q is the mid - point of BP. Then area `(Delta ABQ)` : area `(Delta ABC)` is equal to :

Options

A. `1:3`
B. `1:9`
C. `2:9`
D. `2:5`

Answer

To solve the problem, we need to find the ratio of the area of triangle ABQ to the area of triangle ABC given the conditions in the question.

### Step-by-Step Solution:

1. **Understanding the Ratios**:
- We are given that \( BP:PC = 4:5 \).
- Let \( BP = 4x \) and \( PC = 5x \) for some \( x \).

2. **Finding Lengths**:
- The total length of \( BC \) can be calculated as:
\[ BC = BP + PC = 4x + 5x = 9x \]

3. **Identifying Midpoint Q**:
- Since \( Q \) is the midpoint of \( BP \), we have:
\[ BQ = QP = \frac{BP}{2} = \frac{4x}{2} = 2x \]

4. **Area of Triangle ABQ**:
- The area of triangle \( ABQ \) can be calculated using the formula for the area of a triangle:
\[ \text{Area of } \triangle ABQ = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times BQ \times h \]
- Here, \( BQ = 2x \) and \( h \) is the height from \( A \) to line \( BC \).

5. **Area of Triangle ABC**:
- Similarly, the area of triangle \( ABC \) is:
\[ \text{Area of } \triangle ABC = \frac{1}{2} \times BC \times h = \frac{1}{2} \times 9x \times h \]

6. **Calculating the Ratio**:
- Now, we can find the ratio of the areas:
\[ \frac{\text{Area of } \triangle ABQ}{\text{Area of } \triangle ABC} = \frac{\frac{1}{2} \times 2x \times h}{\frac{1}{2} \times 9x \times h} \]
- The \( \frac{1}{2} \) and \( h \) cancel out:
\[ = \frac{2x}{9x} = \frac{2}{9} \]

7. **Final Result**:
- Therefore, the ratio of the area of triangle \( ABQ \) to the area of triangle \( ABC \) is:
\[ \frac{\text{Area of } \triangle ABQ}{\text{Area of } \triangle ABC} = \frac{2}{9} \]

### Conclusion:
The answer is \( \frac{2}{9} \).