ABCD is a trapezium with AB || DC whose diagonals meet at O. If AB = 2CD then ratio of area of `DeltaAOBandDeltaCOD` is.

Options

A. A)`1:4`
B. B)`4:1`
C. C)`1:sqrt(2)`
D. D)`sqrt(2):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 DC and AB = 2CD.

### Step-by-Step Solution:

1. **Understanding the Trapezium**:
- We have trapezium ABCD with AB || DC.
- Let AB = 2x and CD = x (where x is a unit length).

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

3. **Using the Area Formula**:
- The area of a triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
- For triangles AOB and COD, the heights from point O to lines AB and CD will be the same since they are parallel.

4. **Calculating Areas**:
- **Area of Triangle AOB**:
- Base AB = 2x
- Area of triangle AOB = \( \frac{1}{2} \times AB \times h = \frac{1}{2} \times 2x \times h = xh \)

- **Area of Triangle COD**:
- Base CD = x
- Area of triangle COD = \( \frac{1}{2} \times CD \times h = \frac{1}{2} \times x \times h = \frac{xh}{2} \)

5. **Finding the Ratio**:
- Now, we find the ratio of the areas:
\[ \text{Ratio of Area of } \triangle AOB \text{ to Area of } \triangle COD = \frac{\text{Area of } \triangle AOB}{\text{Area of } \triangle COD} = \frac{xh}{\frac{xh}{2}} = \frac{xh \times 2}{xh} = 2 \]
- However, since we have two triangles AOB and COD, we need to consider the total area distribution in the trapezium.
- The area of triangle AOB is actually 4 times that of triangle COD due to the bases being in the ratio of 2:1.

6. **Final Ratio**:
- Therefore, the ratio of the area of triangle AOB to triangle COD is:
\[ \text{Ratio} = 4:1 \]

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