n equidistant points `A_(1), A_(2), A_(3) ………A_(n)` are taken on base BC of `DeltaABC` is `A_(1)B=A_(1)A_(2)=A_(n)C` and area of `DeltaA""A_(4)A_(5)` is k `cm^(2)`, then what is the area of `DeltaABC`.

Answer

To solve the problem, we need to find the area of triangle ABC given that the area of triangle A_A4_A5 is k cm² and that there are n equidistant points on the base BC of triangle ABC.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have triangle ABC with base BC.
- There are n equidistant points A1, A2, A3, ..., An on base BC.
- The distances are such that A1B = A1A2 = ... = A(n-1)An = AnC.

2. **Dividing Triangle ABC**:
- The points A1, A2, ..., An divide the base BC into n equal segments.
- Each segment has the same height from point A to the base BC.

3. **Area of Triangle**:
- The area of triangle ABC can be expressed in terms of the area of smaller triangles formed by these points.
- The area of triangle A_A4_A5 is given as k cm².

4. **Finding the Area of Smaller Triangles**:
- Since the points A1, A2, ..., An are equidistant, each smaller triangle A_Ai_A(i+1) (for i = 1 to n-1) will have the same height from point A to the base BC.
- The area of each triangle A_Ai_A(i+1) can be expressed as a fraction of the area of triangle ABC.

5. **Calculating the Number of Smaller Triangles**:
- There are (n-1) smaller triangles formed by the points A1 to An.
- The area of each triangle A_Ai_A(i+1) is equal to the total area of triangle ABC divided by n.

6. **Relating Areas**:
- If the area of triangle A_A4_A5 is k cm², then:
\[ \text{Area of triangle ABC} = n \times \text{Area of triangle A_A4_A5} \]
- Therefore, we can substitute k for the area of triangle A_A4_A5:
\[ \text{Area of triangle ABC} = n \times k \]

7. **Final Result**:
- Hence, the area of triangle ABC is:
\[ \text{Area of triangle ABC} = nk \text{ cm}^2 \]

### Final Answer:
The area of triangle ABC is \( nk \) cm².