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?
- A16
- B18
- C20Correct
- D24
Explanation
To find the number of distinct triangles, follow these steps:
-
Identify the total number of points: The problem states there is 1 point on AB, 1 point on CD, 2 points on BC, and 2 points on DA. This gives a total of 1 plus 1 plus 2 plus 2 which equals 6 points.
-
Calculate total possible combinations of three points: The number of ways to choose 3 points out of 6 is calculated as 6C3. 6C3 = (6 x 5 x 4) / (3 x 2 x 1) = 20.
-
Check for collinear points: A triangle cannot be formed if all three points lie on the same straight line. In this case, each side of the square is a straight line. Side AB has only 1 point. Side CD has only 1 point. Side BC has 2 points. Side DA has 2 points.
Since no side contains 3 or more points, it is impossible to pick three collinear points from the chosen set.
- Conclusion: Every combination of 3 points will form a valid triangle. Therefore, the total number of triangles is 20.
The correct answer is C.
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