The number of triangles with each side having integral length and the longest side is of 11 units is equal to `k^(2)`, then the value of 'k' is equal to

Answer

To find the number of triangles with integral side lengths where the longest side is 11 units, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Triangle Inequality**:
For any triangle with sides \(a\), \(b\), and \(c\) (where \(c\) is the longest side), the triangle inequality states that:
- \(a + b > c\)
- \(a > 0\)
- \(b > 0\)

Here, we set \(c = 11\). Therefore, we need to satisfy:
- \(a + b > 11\)
- \(a > 0\)
- \(b > 0\)

2. **Setting Constraints for \(a\) and \(b\)**:
Since \(c = 11\) is the longest side, both \(a\) and \(b\) must be less than or equal to 11:
- \(1 \leq a \leq 11\)
- \(1 \leq b \leq 11\)

3. **Finding Possible Values for \(a\)**:
We will consider each possible value of \(a\) from 1 to 11 and determine the corresponding values of \(b\) such that \(a + b > 11\).

- **If \(a = 11\)**:
\[ b > 11 - 11 \implies b > 0 \quad \text{(possible values: 1 to 11)} \quad \text{(11 values)} \]
- **If \(a = 10\)**:
\[ b > 11 - 10 \implies b > 1 \quad \text{(possible values: 2 to 11)} \quad \text{(10 values)} \]
- **If \(a = 9\)**:
\[ b > 11 - 9 \implies b > 2 \quad \text{(possible values: 3 to 11)} \quad \text{(9 values)} \]
- **If \(a = 8\)**:
\[ b > 11 - 8 \implies b > 3 \quad \text{(possible values: 4 to 11)} \quad \text{(8 values)} \]
- **If \(a = 7\)**:
\[ b > 11 - 7 \implies b > 4 \quad \text{(possible values: 5 to 11)} \quad \text{(7 values)} \]
- **If \(a = 6\)**:
\[ b > 11 - 6 \implies b > 5 \quad \text{(possible values: 6 to 11)} \quad \text{(6 values)} \]
- **If \(a = 5\)**:
\[ b > 11 - 5 \implies b > 6 \quad \text{(possible values: 7 to 11)} \quad \text{(5 values)} \]
- **If \(a = 4\)**:
\[ b > 11 - 4 \implies b > 7 \quad \text{(possible values: 8 to 11)} \quad \text{(4 values)} \]
- **If \(a = 3\)**:
\[ b > 11 - 3 \implies b > 8 \quad \text{(possible values: 9 to 11)} \quad \text{(3 values)} \]
- **If \(a = 2\)**:
\[ b > 11 - 2 \implies b > 9 \quad \text{(possible values: 10 to 11)} \quad \text{(2 values)} \]
- **If \(a = 1\)**:
\[ b > 11 - 1 \implies b > 10 \quad \text{(possible value: 11)} \quad \text{(1 value)} \]

4. **Calculating Total Combinations**:
Now, we sum all the possible values of \(b\) for each corresponding \(a\):
\[ 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 66 \]

5. **Finding \(k\)**:
We are given that the number of triangles is equal to \(k^2\):
\[ k^2 = 66 \]
Thus, \(k = \sqrt{66}\).

### Final Answer:
The value of \(k\) is \( \sqrt{66} \).