The diagram shows triangle ABC with points chosen on each of the sides - HSC - SSCE Mathematics Extension 1 - Question 7 - 2022 - Paper 1

To form a triangle, we must choose 3 distinct points from these 12 available points. The number of ways to choose 3 points from 12 can be calculated using the combination formula:

$C(n, k) = \frac{n!}{k!(n-k)!}$

For our case, this becomes:

$C(12, 3) = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!}$

Calculating the combination gives:

$C(12, 3) = \frac{12\times11\times10}{3\times2\times1} = 220$

From the calculations, we find that the total number of triangles that can be formed is 220.