In `Delta ABC` there are 2 points D and E on BC such that `BD: DE: EC =3:4:5`, if the area of `Delta ADE=40cm^(2)` then find the area of `Delta ABC` .

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Ratios**:
We are given that the segments on line BC are in the ratio \( BD: DE: EC = 3:4:5 \). Let's denote:
- \( BD = 3x \)
- \( DE = 4x \)
- \( EC = 5x \)

2. **Finding the Total Length of BC**:
The total length of \( BC \) can be calculated as:
\[ BC = BD + DE + EC = 3x + 4x + 5x = 12x \]

3. **Area of Triangle ADE**:
We know that the area of triangle \( ADE \) is given as \( 40 \, \text{cm}^2 \).

4. **Calculating the Area of Triangle ADE**:
The area of triangle \( ADE \) can be expressed as:
\[ \text{Area}_{ADE} = \frac{1}{2} \times DE \times h \]
where \( h \) is the height from point \( A \) to line \( BC \). Since \( DE = 4x \), we have:
\[ \text{Area}_{ADE} = \frac{1}{2} \times 4x \times h = 2xh \]
Setting this equal to the given area:
\[ 2xh = 40 \implies xh = 20 \quad \text{(Equation 1)} \]

5. **Calculating the Area of Triangle ABC**:
The area of triangle \( ABC \) can be expressed as:
\[ \text{Area}_{ABC} = \frac{1}{2} \times BC \times h = \frac{1}{2} \times 12x \times h = 6xh \]
Substituting \( xh \) from Equation 1:
\[ \text{Area}_{ABC} = 6 \times 20 = 120 \, \text{cm}^2 \]

6. **Final Answer**:
Therefore, the area of triangle \( ABC \) is:
\[ \text{Area}_{ABC} = 120 \, \text{cm}^2 \]