ABCD is a trapezium in which `AB||CD` and `AB=2CD`. It its diagonals intersect each other at O then ratio of the area of triangle AOB and COD is

Options

A. `1:2`
B. `2:1`
C. `1:4`
D. `4:1`

Answer

To solve the problem, we need to find the ratio of the areas of triangles AOB and COD in trapezium ABCD, where AB is parallel to CD and AB = 2CD.

### Step-by-Step Solution:

1. **Identify the trapezium and its properties**:
- We have a trapezium ABCD where AB || CD.
- Given that AB = 2CD, we can denote CD as x. Therefore, AB = 2x.

2. **Draw the diagonals**:
- Draw diagonals AC and BD, which intersect at point O.

3. **Use properties of parallel lines**:
- Since AB || CD, we know that the angles formed by the transversal (the diagonals) are equal.
- Thus, ∠OAB = ∠OCD (alternate interior angles).
- Similarly, ∠OBA = ∠ODC (alternate interior angles).

4. **Establish angle relationships**:
- The angles ∠AOB and ∠COD are vertically opposite angles, so ∠AOB = ∠COD.

5. **Triangles similarity**:
- From the angle relationships established, we can conclude that triangles AOB and COD are similar by the AA (Angle-Angle) similarity criterion.

6. **Set up the ratio of the areas**:
- The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
- Therefore, we can write:
\[ \frac{\text{Area of } \triangle AOB}{\text{Area of } \triangle COD} = \left(\frac{AB}{CD}\right)^2 \]

7. **Substitute the values**:
- We know that AB = 2CD, so substituting gives:
\[ \frac{\text{Area of } \triangle AOB}{\text{Area of } \triangle COD} = \left(\frac{2CD}{CD}\right)^2 = (2)^2 = 4 \]

8. **Conclusion**:
- Thus, the ratio of the area of triangle AOB to the area of triangle COD is 4:1.

### Final Answer:
The ratio of the area of triangle AOB to the area of triangle COD is **4:1**.