Question

$ABCD$ is a convex quadrilateral and $3,4,5$ and 6 points are marked on the sides $AB,BC,CD$ and $DA,$ respectively. The number of triangles with vertices on different sides is (none of them is $A,B,C$ or$D)$

• A. $270$
• B. $220$
• C. $282$
• D. $342$

Answer 1

Answer: (D)

$342$

Let side A has 3 points on it, side B has 4 points, side C has 5 points and side D has 6 points.

Triangle can be formed by choosing 3 points from different sides.

Since all points should be on different sides so there 4 possible ways of choosing 3 side out of 4 sides.

With side A,B,C we have $3\times 4\times 5=60$ triangles.

With side A,B,D we have $3\times 4\times 6=72$ triangles

With side A,C,D we have $3\times 5\times 6=90$ triangles

With side B,C,D we have $4\times 5\times 6=120$ triangles

Total number of triangles $=60+72+90+120=342$ triangles.