It will be 70.. but I am thinking how to type my scratch work here

To explain it

L1 has 5 points (a,b,c,d,e)
L2 has 4 points (w,x,y,z)

Consider point a on L1 and w on L2 - there are 4 triangles awb, awc, awd, awe. Similarly there will be 4 triangles for each point on L2

Now, think of this, there would be 6 more traingles for point a awx, awy, awz, axy, axz, ayz

In a nutshell No of traingles from each point on L1

= [(no of points on L2)*(no of remaining points on L1)] + 6

So no of triangles
from a = 22
from b = 18
from c = 14
from d = 10
from e = 6

Total = 70

(4 * 5C2) + (5 * 4C2) = 4*10 + 5*6 = 70

givemeanid wrote:(4 * 5C2) + (5 * 4C2) = 4*10 + 5*6 = 70

Thanks givemeanid

Thanks Givemeanid & Kajcha!!!!