Find the values of 'k' if area of the triangle is `4` square unts and vertices are `(2,0),(0,4),(0,k)`.

Answer

To find the values of 'k' such that the area of the triangle formed by the vertices (2,0), (0,4), and (0,k) is 4 square units, we can use the formula for the area of a triangle given its vertices.

### Step-by-step Solution:

1. **Identify the vertices of the triangle**:
The vertices are given as \( A(2, 0) \), \( B(0, 4) \), and \( C(0, k) \).

2. **Use the area formula for a triangle**:
The area \( A \) of a triangle with vertices at \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

3. **Substitute the coordinates into the area formula**:
Here, \( (x_1, y_1) = (2, 0) \), \( (x_2, y_2) = (0, 4) \), and \( (x_3, y_3) = (0, k) \). Plugging in these values, we get:
\[ A = \frac{1}{2} \left| 2(4 - k) + 0(k - 0) + 0(0 - 4) \right| \]
Simplifying this gives:
\[ A = \frac{1}{2} \left| 8 - 2k \right| \]

4. **Set the area equal to 4 square units**:
According to the problem, the area is 4 square units. Therefore, we set up the equation:
\[ \frac{1}{2} \left| 8 - 2k \right| = 4 \]

5. **Multiply both sides by 2 to eliminate the fraction**:
\[ \left| 8 - 2k \right| = 8 \]

6. **Solve the absolute value equation**:
This gives us two cases to consider:
- Case 1: \( 8 - 2k = 8 \)
- Case 2: \( 8 - 2k = -8 \)

**Case 1**:
\[ 8 - 2k = 8 \implies -2k = 0 \implies k = 0 \]

**Case 2**:
\[ 8 - 2k = -8 \implies -2k = -16 \implies k = 8 \]

7. **Conclusion**:
The values of \( k \) that satisfy the condition are \( k = 0 \) and \( k = 8 \).

### Final Answer:
The values of \( k \) are \( 0 \) and \( 8 \).