The sides of a `Delta` are consecutive intergers and inradius is 4 cm. Find the area of the `Delta` .

Answer

To solve the problem, we need to find the area of a triangle with consecutive integer sides and an inradius of 4 cm. Let's denote the sides of the triangle as \(x-1\), \(x\), and \(x+1\).

### Step-by-Step Solution:

1. **Identify the sides of the triangle**:
Let the sides of the triangle be:
- Side 1: \(x - 1\)
- Side 2: \(x\)
- Side 3: \(x + 1\)

2. **Calculate the perimeter of the triangle**:
The perimeter \(P\) of the triangle is the sum of its sides:
\[ P = (x - 1) + x + (x + 1) = 3x \]

3. **Use the formula for the area of the triangle**:
The area \(A\) of a triangle can also be calculated using the formula:
\[ A = \frac{1}{2} \times r \times P \]
where \(r\) is the inradius and \(P\) is the perimeter. Given that the inradius \(r = 4\) cm, we can substitute the values:
\[ A = \frac{1}{2} \times 4 \times (3x) = 6x \]

4. **Find the value of \(x\)**:
To find \(x\), we need to ensure that the triangle inequality holds for the sides \(x-1\), \(x\), and \(x+1\):
- \( (x - 1) + x > (x + 1) \) simplifies to \(2x - 1 > x + 1\) or \(x > 2\).
- \( (x - 1) + (x + 1) > x \) simplifies to \(2x > x\) which is always true.
- \( x + (x + 1) > (x - 1) \) simplifies to \(2x + 1 > x - 1\) or \(x > -2\) which is always true.

The critical condition is \(x > 2\). Since \(x\) must be an integer, the smallest value for \(x\) is 3.

5. **Calculate the area**:
Substitute \(x = 3\) into the area formula:
\[ A = 6x = 6 \times 3 = 18 \text{ cm}^2 \]

### Final Answer:
The area of the triangle is \(18 \text{ cm}^2\).