Q20. ABCD is a square. One point on each of AB and CD; and two distinct points on each of BC and DA are chosen. How many distinct triangles can be drawn using any three points as vertices out of these six points?

Let's list the chosen points: - One point on side AB (let's call it P1) - One point on side CD (P2) - Two distinct points on side BC (P3, P4) - Two distinct points on side DA (P5, P6) In total, there are 1 + 1 + 2 + 2 = 6 distinct points. To form a triangle, we need to choose any 3 points from these 6 points.

The total number of ways to choose 3 points from 6 is given by the combination formula 6C3. 6C3 = 6! / (3! * (6-3)!) = (6 × 5 × 4) / (3 × 2 × 1) = 20. For these points to form a triangle, no three chosen points should be collinear.

In this setup, points P3 and P4 are collinear (on BC), and P5 and P6 are collinear (on DA). However, no three points from the entire set of 6 can be collinear. For example, P1 (on AB), P3 (on BC), and P5 (on DA) cannot be collinear.

Similarly, P1, P3, P4 cannot be collinear as P1 is on AB and P3, P4 are on BC. Since no three points are collinear, every combination of 3 points will form a distinct triangle. Thus, the total number of distinct triangles is 20.