In `Delta ABC`, D and E are two points on the sides AB and AC respectively so that DE || BC and `(AD)/(BD) = (2)/(3)`. Then `("the area of trapezium DECB")/("the area of " Delta ABC "equa to")`

Options

A. `(5)/(9)`
B. `(21)/(25)`
C. `1(4)/(5)`
D. `5(1)/(4)`

Answer

To solve the problem, we need to find the ratio of the area of trapezium DECB to the area of triangle ABC given that DE is parallel to BC and the ratio of segments AD to BD is 2:3.

### Step-by-Step Solution:

1. **Identify the segments**:
- Let AD = 2k and BD = 3k (where k is a positive integer).
- Therefore, AB = AD + BD = 2k + 3k = 5k.

2. **Establish similarity of triangles**:
- Since DE is parallel to BC, triangles ADE and ABC are similar by the Angle-Angle (AA) similarity criterion.
- This means that the corresponding sides of these triangles are proportional.

3. **Set 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.
- Thus, we have:
\[ \frac{\text{Area of } \triangle ADE}{\text{Area of } \triangle ABC} = \left(\frac{AD}{AB}\right)^2 = \left(\frac{2k}{5k}\right)^2 = \left(\frac{2}{5}\right)^2 = \frac{4}{25}. \]

4. **Calculate the area of triangle ABC**:
- Let the area of triangle ABC be denoted as Area(ABC) = 25k² (as derived from the ratio).
- Therefore, Area(ADE) = \(\frac{4}{25} \times \text{Area(ABC)} = \frac{4}{25} \times 25k^2 = 4k^2\).

5. **Find the area of trapezium DECB**:
- The area of trapezium DECB can be calculated as:
\[ \text{Area of trapezium DECB} = \text{Area of triangle ABC} - \text{Area of triangle ADE} = 25k^2 - 4k^2 = 21k^2. \]

6. **Calculate the ratio of the areas**:
- Now, we find the ratio of the area of trapezium DECB to the area of triangle ABC:
\[ \frac{\text{Area of trapezium DECB}}{\text{Area of triangle ABC}} = \frac{21k^2}{25k^2} = \frac{21}{25}. \]

### Final Answer:
The ratio of the area of trapezium DECB to the area of triangle ABC is \(\frac{21}{25}\).