In a `Delta ` ABC , D and E are two points on BC such that BD : DE : EC = 4 : 5 : 6 If the area of `Delta ADE = 30 cm^(2)`. find area of `Delta ABC `

Answer

To find the area of triangle ABC given the area of triangle ADE and the ratios of segments on BC, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Ratios**:
The problem states that the segments on BC are in the ratio \( BD : DE : EC = 4 : 5 : 6 \). Let's denote:
- \( BD = 4k \)
- \( DE = 5k \)
- \( EC = 6k \)

Here, \( k \) is a common factor.

2. **Calculate Total Length of BC**:
The total length of segment BC can be calculated as:
\[ BC = BD + DE + EC = 4k + 5k + 6k = 15k \]

3. **Area Relationships**:
The area of triangles sharing the same height from point A to line BC can be expressed in terms of their bases:
- Area of triangle \( ABD \) is proportional to \( BD \)
- Area of triangle \( ADE \) is proportional to \( DE \)
- Area of triangle \( AEC \) is proportional to \( EC \)

Thus, we can express the areas as:
- Area of triangle \( ABD = 4x \)
- Area of triangle \( ADE = 5x \)
- Area of triangle \( AEC = 6x \)

4. **Given Area of Triangle ADE**:
We know from the problem that:
\[ \text{Area of triangle } ADE = 30 \, \text{cm}^2 \]
Therefore, we can set:
\[ 5x = 30 \implies x = \frac{30}{5} = 6 \, \text{cm}^2 \]

5. **Calculate Areas of Other Triangles**:
Now we can find the areas of triangles \( ABD \) and \( AEC \):
- Area of triangle \( ABD = 4x = 4 \times 6 = 24 \, \text{cm}^2 \)
- Area of triangle \( AEC = 6x = 6 \times 6 = 36 \, \text{cm}^2 \)

6. **Calculate Total Area of Triangle ABC**:
The total area of triangle ABC is the sum of the areas of triangles \( ABD \), \( ADE \), and \( AEC \):
\[ \text{Area of triangle } ABC = \text{Area of } ABD + \text{Area of } ADE + \text{Area of } AEC \]
\[ = 24 + 30 + 36 = 90 \, \text{cm}^2 \]

### Final Answer:
The area of triangle ABC is \( 90 \, \text{cm}^2 \).

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