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# PS: Number of triangles...

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Manager
Joined: 17 Dec 2008
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21 Feb 2009, 19:57
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1. ABCDE is a regular pentagon with F the center. How many different points can be formed by joining 3 of the points?

A. 10
B. 20
C. 120
D. 5
E. 6
SVP
Joined: 07 Nov 2007
Posts: 1761
Location: New York
Re: PS: Number of triangles... [#permalink]

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21 Feb 2009, 23:07
ConkergMat wrote:
1. ABCDE is a regular pentagon with F the center. How many different points can be formed by joining 3 of the points?

A. 10
B. 20
C. 120
D. 5
E. 6

joining 3 points --> results TRIANGLE not POINT

I am assuming How many different triangles can be formed?

6C3 = 6*5*4 /3! = 20

B.
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Manager
Joined: 17 Dec 2008
Posts: 166
Re: PS: Number of triangles... [#permalink]

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22 Feb 2009, 08:35
Doesn't every triangle have to include F the center point?
Hence shouldn't the answer be 5C2 = 10?
Manager
Joined: 27 May 2008
Posts: 198
Re: PS: Number of triangles... [#permalink]

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22 Feb 2009, 09:18
5C2 = 10

Since it is pentagon, we can connect any point of the pentagon to the opposite two pints.
VP
Joined: 05 Jul 2008
Posts: 1373
Re: PS: Number of triangles... [#permalink]

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22 Feb 2009, 14:22
ConkergMat wrote:
Doesn't every triangle have to include F the center point?
Hence shouldn't the answer be 5C2 = 10?

Not necessarily, The line that connects the points that form the rectangle/square piece of the pentagon does not need to go throug F. Similarly many obtuse triangles can be made without passing through the center. Q comes down to 6 points, how many triangles??
Manager
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Re: PS: Number of triangles... [#permalink]

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09 Mar 2009, 10:12
x2suresh wrote:
ConkergMat wrote:
1. ABCDE is a regular pentagon with F the center. How many different points can be formed by joining 3 of the points?

A. 10
B. 20
C. 120
D. 5
E. 6

joining 3 points --> results TRIANGLE not POINT

I am assuming How many different triangles can be formed?

6C3 = 6*5*4 /3! = 20

B.

Can You explain how You came to 6C3 as the answer,

According to what i understand, to form a triangle with any three vertices of the pentagon you will atleast have one of the sides of the pentagon as a side of your triangle.

And with each side if you try to draw a triangle using the other vertices you can draw a maximum of 3 triangles

since it's a pentagon we have 3*5 = 15 triangles. But that's not one of the options.. Could some one explain where i am going wrong?
Director
Joined: 25 Oct 2006
Posts: 608
Re: PS: Number of triangles... [#permalink]

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09 Mar 2009, 11:29

You have total 6 points and select any 3 points out of 6. Hence 6C3 = 20 is perfect. Now see the pictorial presentation.
Attachment:

triangle.JPG [ 3.12 KiB | Viewed 992 times ]

With each side 4 triangles are possible so with 5 sides 5*4=20 triangles are possible.
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Re: PS: Number of triangles...   [#permalink] 09 Mar 2009, 11:29
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