To solve this problem, we need to determine the number of triangles with integer side lengths where the longest side is 11. For a valid triangle, the sum of the lengths of any two sides must be greater than the length of the third side.

Let's denote the sides of the triangle as $a$, $b$, and $c$, where $c = 11$ (the longest side), and $a$ and $b$ are the other two sides with $a \leq b \leq c$. The triangle inequality gives us the following conditions:

1. $a + b > c$
2. $a + c > b$
3. $b + c > a$

Since $c = 11$, the conditions simplify to:

1. $a + b > 11$
2. $a + 11 > b$
3. $b + 11 > a$

For $a + b > 11$, and given that $a \leq b$, we need to consider the range for $a$ and $b$.

Case Analysis:

We analyze for different values of $a$ and find the corresponding values of $b$:

• $a = 1$: $1 + b > 11$ implies $b > 10$, so $b = 11$.
• $a = 2$: $2 + b > 11$ implies $b > 9$, so $b = 10, 11$.
• $a = 3$: $3 + b > 11$ implies $b > 8$, so $b = 9, 10, 11$.
• $a = 4$: $4 + b > 11$ implies $b > 7$, so $b = 8, 9, 10, 11$.
• $a = 5$: $5 + b > 11$ implies $b > 6$, so $b = 7, 8, 9, 10, 11$.
• $a = 6$: $6 + b > 11$ implies $b > 5$, so $b = 6, 7, 8, 9, 10, 11$.
• $a = 7$: $7 + b > 11$ implies $b > 4$, so $b = 5, 6, 7, 8, 9, 10, 11$.
• $a = 8$: $8 + b > 11$ implies $b > 3$, so $b = 4, 5, 6, 7, 8, 9, 10, 11$.
• $a = 9$: $9 + b > 11$ implies $b > 2$, so $b = 3, 4, 5, 6, 7, 8, 9, 10, 11$.
• $a = 10$: $10 + b > 11$ implies $b > 1$, so $b = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11$.
• $a = 11$: $11 + b > 11$ implies $b > 0$, so $b = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11$.

Now, let's count the valid pairs $(a, b)$:

&\text{For } a = 1, \text{ one pair: } (1, 11) \\ &\text{For } a = 2, \text{ two pairs: } (2, 10), (2, 11) \\ &\text{For } a = 3, \text{ three pairs: } (3, 9), (3, 10), (3, 11) \\ &\text{For } a = 4, \text{ four pairs: } (4, 8), (4, 9), (4, 10), (4, 11) \\ &\text{For } a = 5, \text{ five pairs: } (5, 7), (5, 8), (5, 9), (5, 10), (5, 11) \\ &\text{For } a = 6, \text{ six pairs: } (6, 6), (6, 7), (6, 8), (6, 9), (6, 10), (6, 11) \\ &\text{For } a = 7, \text{ six pairs: } (7, 5), (7, 6), (7, 7), (7, 8), (7, 9), (7, 10) \\ &\text{For } a = 8, \text{ four pairs: } (8, 4), (8, 5), (8, 6), (8, 7) \\ &\text{For } a = 9, \text{ three pairs: } (9, 3), (9, 4), (9, 5) \\ &\text{For } a = 10, \text{ two pairs: } (10, 2), (10, 3) \\ &\text{For } a = 11, \text{ one pair: } (11, 1) \\ \end{aligned}$$ ### Total Number of Triangles: Adding all the valid pairs gives us: $$1 + 2 + 3 + 4 + 5 + 6 + 6 + 4 + 3 + 2 + 1 = 37$$ So, there are **37** triangles where each side has an integral length, and the longest side is 11. Would you like more details or have any other questions? --- **Related Questions:** 1. How many triangles can be formed if the longest side is 12? 2. What is the sum of all possible perimeters of triangles with the longest side being 11? 3. How does the triangle inequality affect the possible side lengths? 4. Can the above method be generalized to any given longest side? 5. How many obtuse triangles are there with a maximum side length of 11? **Tip:** When counting valid triangles, always ensure the triangle inequality holds for all sides to avoid counting impossible triangles.