Question

If pqrs is a convex quadrilateral with 3,4,5 and 6 points marked on sides pq,qr,rs and ps respectively then find the number of triangle with vertices on different sides

Answer 1

just marks the points or draw diagnols to find the triangle acc to me there are 16

Answer 2

Answer: 342

Given:    ∆XYZ ~∆LMN (i.e. they are similar)

Solution:

Concept: C(n.r)= $\frac { { n! } }{ r!(n-r)! }$

The maximum number of triangles can be formed with vertices on side AB, BC, $CD=3C_{1}\times 4C_{1}\times 5C_{1}$

The same for other combinations: BC, CD, DA and CD, DA, AB and DA, AB, BC

Totally 4 combinations

Hence, the total number of triangles can be obtained by the total sum:

           $= (3C_{1}\times4C_{1}\times5C_{1}) + (3C_{1}\times4C_{1}\times6C_{1}) + (3C_{1}\times5C_{1}\times6C_{1}) + (4C_{1}\times5C_{1}\times6C_{1})$

$3C_{1}=  \frac{{3!}} {1!2!} = 3\\4C_{1}=  \frac{{4!}} {1!3!}  = 4\\5C_{1}=  \frac{{5!}} {(1!4!)}  = 5\\6C_{1}= \frac{{ 6!}} {1!5!}  = 6$

         ⇒$(3 \times 4 \times 5) + (3 \times 4 \times 6) + (3 \times 5 \times 6) + (4 \times 5 \times 6) = 342$