In a trapezium ABCD, O is the point of intesection of AC and BD, AB||CD and `AB = 2xx CD`. If the area of `triangleAOB= 84cm^(2)` . Find the area of `triangleCOD`.

Options

A. `42cm^(2)`
B. `21cm^(2)`
C. `25cm^(2)`
D. `40cm^(2)`

Explanation

To solve the problem step by step, we will use the properties of similar triangles and the relationship between their areas.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- In trapezium ABCD, AB || CD.
- AB = 2 * CD.
- Area of triangle AOB = 84 cm².
- We need to find the area of triangle COD.

2. **Understanding the Similar Triangles:**
- Since AB is parallel to CD, the angles formed by the intersection of the diagonals AC and BD create alternate angles.
- Therefore, we have:
- ∠AOB = ∠COD (alternate angles)
- ∠OAB = ∠OCD (alternate angles)
- This means that triangles AOB and COD are similar by the AA (Angle-Angle) criterion.

3. **Setting Up the Ratio of Areas:**
- For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
- Let the length of CD be x. Then, the length of AB is 2x.
- The ratio of the lengths of AB to CD is:
\[ \frac{AB}{CD} = \frac{2x}{x} = 2 \]
- The ratio of the areas of triangles AOB and COD is:
\[ \frac{\text{Area of } \triangle AOB}{\text{Area of } \triangle COD} = \left(\frac{AB}{CD}\right)^2 = 2^2 = 4 \]

4. **Using the Area of Triangle AOB:**
- Let the area of triangle COD be denoted as A.
- From the ratio of the areas, we have:
\[ \frac{84}{A} = 4 \]
- Cross-multiplying gives:
\[ 84 = 4A \]

5. **Solving for A:**
- Dividing both sides by 4:
\[ A = \frac{84}{4} = 21 \text{ cm}^2 \]

6. **Conclusion:**
- The area of triangle COD is 21 cm².

### Final Answer:
The area of triangle COD is **21 cm²**.