In a trapezium ABCD, the diagonals AC and BD interserted other at O such that `OB : OD = 3:1` then the ratio of area of `DeltaAOB: DeltaCOD` is :

Options

A. `3:1`
B. `1:4`
C. `9:1`
D. can't be determined

Answer

To solve the problem, we need to find the ratio of the areas of triangles AOB and COD in trapezium ABCD, given that the diagonals AC and BD intersect at point O such that the ratio OB:OD = 3:1.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have a trapezium ABCD with parallel sides AB and CD.
- The diagonals AC and BD intersect at point O.

2. **Given Ratio**:
- We know that OB:OD = 3:1. This means that if we let OD = 1 unit, then OB = 3 units.

3. **Using the Properties of Similar Triangles**:
- Since AB is parallel to CD, triangles AOB and COD are similar by the Alternate Interior Angles Theorem.
- Therefore, the ratio of their areas is proportional to the square of the ratio of their corresponding sides.

4. **Finding the Ratio of Sides**:
- The sides corresponding to triangles AOB and COD are OB and OD, respectively.
- Thus, the ratio of the sides is given by:
\[ \frac{OB}{OD} = \frac{3}{1} \]

5. **Calculating the Ratio of Areas**:
- The ratio of the areas of similar triangles is the square of the ratio of their corresponding sides:
\[ \frac{\text{Area of } AOB}{\text{Area of } COD} = \left(\frac{OB}{OD}\right)^2 = \left(\frac{3}{1}\right)^2 = \frac{9}{1} \]

6. **Final Result**:
- Therefore, the ratio of the area of triangle AOB to the area of triangle COD is:
\[ \text{Area of } AOB : \text{Area of } COD = 9 : 1 \]

### Conclusion:
The ratio of the area of triangle AOB to the area of triangle COD is **9:1**.