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 solve the problem of finding the number of triangles with integral side lengths where the longest side is 11 units, we can use the triangle inequality theorem. The steps are as follows:

### Step 1: Understanding the Triangle Inequality
For any triangle with sides \( a \), \( b \), and \( c \) (where \( c \) is the longest side), the triangle inequality states:
1. \( a + b > c \)
2. \( a + c > b \)
3. \( b + c > a \)

In this case, since the longest side \( c = 11 \), we need to ensure that \( a + b > 11 \).

### Step 2: Setting Up the Problem
Let \( a \) and \( b \) be the other two sides of the triangle. According to the triangle inequality:
- \( a + b > 11 \)
- \( a \) and \( b \) must be positive integers.

### Step 3: Finding Possible Values for \( a \) and \( b \)
Since \( c = 11 \) is the longest side, both \( a \) and \( b \) must be less than or equal to 11. Therefore, we can express the conditions for \( a \) and \( b \) as:
- \( 1 \leq a \leq 11 \)
- \( 1 \leq b \leq 11 \)

### Step 4: Counting Valid Combinations
We can count the valid pairs \( (a, b) \) such that \( a + b > 11 \):
1. If \( a = 11 \), then \( b \) can be from 1 to 11. This gives us 11 combinations.
2. If \( a = 10 \), then \( b \) must be greater than 1 (i.e., \( b = 2, 3, \ldots, 10 \)). This gives us 9 combinations.
3. If \( a = 9 \), then \( b \) must be greater than 2 (i.e., \( b = 3, 4, \ldots, 9 \)). This gives us 7 combinations.
4. If \( a = 8 \), then \( b \) must be greater than 3 (i.e., \( b = 4, 5, \ldots, 8 \)). This gives us 5 combinations.
5. If \( a = 7 \), then \( b \) must be greater than 4 (i.e., \( b = 5, 6, 7 \)). This gives us 3 combinations.
6. If \( a = 6 \), then \( b \) must be greater than 5 (i.e., \( b = 6 \)). This gives us 1 combination.

### Step 5: Summing the Combinations
Now, we sum all the valid combinations:
- For \( a = 11 \): 11 combinations
- For \( a = 10 \): 9 combinations
- For \( a = 9 \): 7 combinations
- For \( a = 8 \): 5 combinations
- For \( a = 7 \): 3 combinations
- For \( a = 6 \): 1 combination

Total = \( 11 + 9 + 7 + 5 + 3 + 1 = 36 \)

### Step 6: Relating to \( k^2 \)
According to the problem, the total number of triangles is equal to \( k^2 \). We found that the total number of triangles is 36, so:
\[ k^2 = 36 \]
Taking the square root of both sides, we find:
\[ k = 6 \]

### Final Answer
The value of \( k \) is \( 6 \).

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