ABCD is a trapezium in which AB||CD and AB=2CD. If its diagonals intersect each other at O, then ratio of areas of trianlges AOB and COD is :

Options

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

Answer

To find the ratio of the areas of triangles AOB and COD in trapezium ABCD where AB is parallel to CD and AB = 2CD, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the trapezium configuration**:
- Let AB = 2x and CD = x (since AB is twice CD).
- The trapezium ABCD has parallel sides AB and CD.

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

3. **Identify the triangles**:
- We need to find the areas of triangles AOB and COD.

4. **Use the properties of similar triangles**:
- Since AB || CD, the angles formed by the diagonals with the parallel sides are equal:
- ∠AOB = ∠COD (alternate interior angles)
- ∠OAB = ∠OCD (alternate interior angles)

5. **Establish the ratio of the bases**:
- The base of triangle AOB is AB = 2x.
- The base of triangle COD is CD = x.

6. **Use the formula for the area of triangles**:
- The area of a triangle is given by the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
- Since both triangles share the same height from point O to line AB and CD, we can express the areas as:
- Area of triangle AOB = \(\frac{1}{2} \times (2x) \times h\)
- Area of triangle COD = \(\frac{1}{2} \times x \times h\)

7. **Calculate the areas**:
- Area of triangle AOB = \(x \times h\)
- Area of triangle COD = \(\frac{1}{2} \times x \times h\)

8. **Find the ratio of the areas**:
- The ratio of the areas of triangles AOB and COD is:
\[ \frac{\text{Area of AOB}}{\text{Area of COD}} = \frac{x \times h}{\frac{1}{2} \times x \times h} = \frac{x \times h}{\frac{1}{2} x \times h} = \frac{2}{1} \]

9. **Final ratio**:
- Therefore, the ratio of the areas of triangles AOB and COD is 4:1.