Diagonals of a Trapezium `DeltaBCD` with AB || CD intersect each other at the point O. If AB = 2CD, then the ratio of the areas of `Delta AOB` and `Delta COD` is

Options

A. `4:1`
B. `1:16`
C. `1:4`
D. `16:1`

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. **Understanding the Trapezium**:
- We have a trapezium ABCD with AB || CD. This means that the sides AB and CD are parallel to each other.

2. **Identifying the Given Information**:
- We are given that AB = 2CD. Let’s denote CD as x. Therefore, AB = 2x.

3. **Analyzing the Triangles**:
- The diagonals AC and BD intersect at point O. We need to find the areas of triangles AOB and COD.

4. **Using Properties of Similar Triangles**:
- Since AB || CD, we can use the properties of similar triangles. The angles formed by the intersection of the diagonals and the parallel sides lead to the conclusion that:
- Angle AOB = Angle COD (alternate interior angles)
- Angle OAB = Angle OCD (alternate interior angles)
- Thus, triangles AOB and COD are similar.

5. **Setting Up the Ratio of Areas**:
- The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore:
\[ \frac{\text{Area of } \triangle AOB}{\text{Area of } \triangle COD} = \left(\frac{AB}{CD}\right)^2 \]

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

7. **Final Ratio**:
- Therefore, the ratio of the areas of triangles AOB and COD is:
\[ \frac{\text{Area of } \triangle AOB}{\text{Area of } \triangle COD} = 4:1 \]

### Conclusion:
The ratio of the areas of triangle AOB to triangle COD is **4:1**.