Question

$ABCD$ is a convex quadrilateral. $3,4,5$ and $6$ points are marked on the sides $AB,BC,CD$ and $DA$ respectively. The no. of triangles with vertices on different sides are?

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

Answer 1

The correct option is D

$342$

Explanation for the correct option:

Find the number of triangles:

Given that there are $3$ points on $AB$, $4$ on $BC$, $5$ on $CD$, $6$ on $DA$

There are $4$ possibilities to find the triangles whose vertices lie on different sides;

1. $AB,BC,CD:(3,4,5)\to {}^{3}C_{1}x{}^{4}C_{1}x{}^{5}C_1=60$
2. $BC,CD,DA:(4,5,6)\to {}^{4}C_{1}x{}^{5}C_{1}x{}^{6}C_1=120$
3. $CD,DA,AB:(5,6,3)\to {}^{5}C_{1}x{}^{6}C_{1}x{}^{3}C_1=90$
4. $DA,AB,BC:(6,3,4)\to {}^{6}C_{1}x{}^{3}C_{1}x{}^{4}C_1=72$

$\therefore$Total number of triangles $=60+120+90+72$

$=342$

Hence, option ‘D’ is correct.