The problem asks for how many different (non-congruent) triangles can be drawn with sides that are integers, where the perimeter is 12 cm.

Solution Outline:

We are looking for integer side lengths $a$, $b$, and $c$ such that:

1. $a + b + c = 12$ (since the perimeter is 12 cm)
2. The sides must satisfy the triangle inequality:
   • $a + b > c$
   • $a + c > b$
   • $b + c > a$

Step-by-Step Solution:

Given the perimeter is 12, we can write the equation:

$a + b + c = 12$ From this, $c = 12 - a - b$. Now, substitute this into the triangle inequality conditions:

1. $a + b > c \Rightarrow a + b > 12 - a - b \Rightarrow 2a + 2b > 12 \Rightarrow a + b > 6$
2. $a + c > b \Rightarrow a + (12 - a - b) > b \Rightarrow 12 - b > b \Rightarrow b < 6$
3. Similarly, $b + c > a \Rightarrow b + (12 - a - b) > a \Rightarrow 12 - a > a \Rightarrow a < 6$

So, we need to find pairs of integers $a$, $b$, and $c = 12 - a - b$ such that $a + b + c = 12$, and $a + b > 6$.

We now check all possible combinations of $a$, $b$, and $c$ under these conditions:

Thus, there are 5 different (non-congruent) triangles.

Answer:

The correct answer is 5.

Would you like further details or explanations?

Related Questions:

1. How can we determine the number of non-congruent triangles for a different perimeter?
2. What are the triangle inequality conditions?
3. Can two triangles with the same perimeter and different side lengths be congruent?
4. What are the properties of integer-sided triangles?
5. How does the triangle inequality relate to different types of triangles?

Tip: Always check the triangle inequality after finding the side lengths to ensure a valid triangle!