Answer: 135.

Solution. To form a triangle, one needs to choose two points on one side and one point on another. We have: 5 ways to choose the first point on $AB$, 4 ways - the second, and since the triangle does not change with the permutation of its vertices, we divide $5 \cdot 4$ by 2. Thus, $\frac{5 \cdot 4}{2}=10$ ways to choose two points on side $AB$, and 6 ways on side $BC$. In total, $10 \cdot 6=60$ ways. Similarly, $\frac{6 \cdot 5}{2}=15$ ways to choose two points on side $BC$ and 5 ways to choose one point on side $AB$. In total, $15 \cdot 5=75$ ways. Therefore, there are 135 different triangles with vertices at the chosen points.
2. (17 points) Given a rectangle $A B C D$. On two sides of the rectangle, different points are chosen, five points on $A B$ and six on $B C$. How many different triangles exist with vertices at the chosen points?