Five points lie on a straight line

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imago: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Sun Sep 09, 2007 3:34 am; Location: spain

Five points lie on a straight line

by imago » Tue Sep 18, 2007 1:19 pm

Five points lie on a straight line and another 4 points lie on another straight line that is parallel to the first one, how many different triangles can be made by linking these 9 points?

a.7
b.20
c.70
d.140
e.400

OA: C

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Source: Beat The GMAT — Problem Solving | Tue Sep 18, 2007 1:19 pm

kajcha: Master | Next Rank: 500 Posts; Posts: 321; Joined: Tue Aug 28, 2007 5:42 am; Thanked: 1 times

by kajcha » Tue Sep 18, 2007 1:30 pm

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

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kajcha: Master | Next Rank: 500 Posts; Posts: 321; Joined: Tue Aug 28, 2007 5:42 am; Thanked: 1 times

by kajcha » Tue Sep 18, 2007 2:24 pm

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

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givemeanid: Master | Next Rank: 500 Posts; Posts: 277; Joined: Sun Jun 17, 2007 2:51 pm; Location: New York, NY; Thanked: 6 times; Followed by:1 members