Question

The number of triangles with each side having integral length and the longest side is of 11 units is equal to k², then the value of k is equal to

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

Answer: (B)

6

Answer 2

Let the sides of the triangle be $a$, $b$, and $c$. We are given that the sides have integral length and the longest side is 11 units. Let the sides be ordered as $s_1 \le s_2 \le s_3$. The condition "longest side is of 11 units" implies $s_3 = 11$. So, we have $1 \le s_1 \le s_2 \le 11$.

The triangle inequality states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. For sides $s_1, s_2, s_3$, these inequalities are:

1. $s_1 + s_2 > s_3$
2. $s_1 + s_3 > s_2$
3. $s_2 + s_3 > s_1$

Substituting $s_3 = 11$:

1. $s_1 + s_2 > 11$
2. $s_1 + 11 > s_2$
3. $s_2 + 11 > s_1$

Since we have the condition $s_1 \le s_2 \le 11$, the inequalities (2) and (3) are automatically satisfied:

• For (2): $s_1 + 11 > s_2$ is true because $s_1 \ge 1$, so $s_1 + 11 \ge 1 + 11 = 12$, and $s_2 \le 11$. Thus, $s_1 + 11 > s_2$.
• For (3): $s_2 + 11 > s_1$ is true because $s_2 \ge s_1$ and $11 > 0$, so $s_2 + 11 > s_1$.

Therefore, we only need to satisfy the conditions:

• $s_1, s_2$ are positive integers.
• $1 \le s_1 \le s_2 \le 11$
• $s_1 + s_2 > 11$

We can find the number of such pairs $(s_1, s_2)$ by iterating through possible values of $s_1$ from 1 to 11:

• If $s_1 = 1$: $s_2 > 10$, so $s_2 = 11$. (1 pair)
• If $s_1 = 2$: $s_2 > 9$, so $s_2 \in \{10, 11\}$. (2 pairs)
• If $s_1 = 3$: $s_2 > 8$, so $s_2 \in \{9, 10, 11\}$. (3 pairs)
• If $s_1 = 4$: $s_2 > 7$, so $s_2 \in \{8, 9, 10, 11\}$. (4 pairs)
• If $s_1 = 5$: $s_2 > 6$, so $s_2 \in \{7, 8, 9, 10, 11\}$. (5 pairs)
• If $s_1 = 6$: $s_2 > 5$. Since $s_2 \ge 6$, all $s_2 \in \{6, 7, 8, 9, 10, 11\}$ satisfy this. (6 pairs)
• If $s_1 = 7$: $s_2 > 4$. Since $s_2 \ge 7$, all $s_2 \in \{7, 8, 9, 10, 11\}$ satisfy this. (5 pairs)
• If $s_1 = 8$: $s_2 > 3$. Since $s_2 \ge 8$, all $s_2 \in \{8, 9, 10, 11\}$ satisfy this. (4 pairs)
• If $s_1 = 9$: $s_2 > 2$. Since $s_2 \ge 9$, all $s_2 \in \{9, 10, 11\}$ satisfy this. (3 pairs)
• If $s_1 = 10$: $s_2 > 1$. Since $s_2 \ge 10$, all $s_2 \in \{10, 11\}$ satisfy this. (2 pairs)
• If $s_1 = 11$: $s_2 > 0$. Since $s_2 = 11$, it satisfies this. (1 pair)

The total number of such pairs $(s_1, s_2)$ is the sum of the counts for each $s_1$: Total number of triangles = $1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 36$.

This number is given to be equal to $k^2$. So, $k^2 = 36$. Taking the square root, $k = \sqrt{36} = 6$.