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Interesting Geometry Question / Triangle

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by pavell » Tue Aug 11, 2009 6:29 am
Interesting question . What is answer?
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by georgeung » Tue Aug 11, 2009 10:06 am
I am going to say Answer B.

I counted in my head.

10 triangles using F.

5 triangles without F.

I'm interested in seeing how to solve this in a more efficient manner.
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by Svedankae » Tue Aug 11, 2009 1:32 pm
well.... heres the thing. the official answer says D) = 20 triangles.


however i have no idea how they come up with that. I am getting 15 triangles but not 20.
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by georgeung » Tue Aug 11, 2009 1:37 pm
LOL. Damn, we're stuck in the water. Hmm, I hope someone comes in here and helps us find the answer.
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There are 20 triangles as follows

by mbadreams » Tue Aug 11, 2009 3:01 pm
1.ABC 11.BCD
2.ABD 12.BCE
3.ABE 13.BCF
4.ABF 14.BDE
5.ACD 15.BDF
6.ACE 16.BEF
7.ACF 17.CDE
8.ADE 18.CDF
9.ADF 19.CEF
10.AEF 20.DEF

didnt find a easier way but to sit and write the different combinations... anyone found an easier way.. please post...
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by prindaroy » Tue Aug 11, 2009 8:45 pm
Okay, so there are 6 points. Now three points together can form a triangle. So we need to choose 3 points from those 6 points.

So the answer is simply 6C3 = 20
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by georgeung » Tue Aug 11, 2009 8:46 pm
Really? It was a combination problem?

Thanks Prindaroy for helping us out here.
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by Svedankae » Wed Aug 12, 2009 3:43 am
ah thanks guys. :)
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by sreak1089 » Wed Aug 12, 2009 3:48 am
You have 5 points of which you can select any two to combine with the third point that is existing at the center to form a triangle, which can be achieved in 5C2 ways == 10.
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by sreak1089 » Wed Aug 12, 2009 3:50 am
Oh no, its 6C3, hence 20 ways.

I thought how many triangles will be formed always joining point F (center of the pentagon).
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by knightwalker » Wed Aug 12, 2009 10:03 pm
yes 6C3=20 is right, but to think of it conceptually for those who are confused, I would say look at it like this:

5 triangles for F as a point (using each of the sides of the pentagon as a base)
plus, 3 triangles for EVERY point of the pentagon (see the figure... eg. for point C, ACE, ECD, and BCA) --- for five points you get 15 triangles

15+5 = 20

I'm still working on my permutations and combinations (my weakest point) so it helps me to figure it out conceptually :)
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by georgeung » Wed Aug 12, 2009 10:10 pm
Okay, say we use the same image above, but remove F. We have Pentagon with points ABCDE.

In order to solve, we now use 5C3, correct?

I guess it makes sense. Say we have square ABCD and ask how many triangles, then we use 4C3, correct?

Thanks in advance. You guys are awesome.
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I agree with 6C3.

by struggling_guy2001 » Wed Aug 12, 2009 10:18 pm
The question is how many traingles can be made with the help of points A,B,C,D,E and F.

F is one among the points and not the compulsory one.

Hence , there are 6 points and we need to choose 3 points to make a triangle.

Hence the answer would be 6C3= 20.


Hope it is clear...
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by knightwalker » Wed Aug 12, 2009 11:20 pm
georgeung wrote:Okay, say we use the same image above, but remove F. We have Pentagon with points ABCDE.

In order to solve, we now use 5C3, correct?

I guess it makes sense. Say we have square ABCD and ask how many triangles, then we use 4C3, correct?

Thanks in advance. You guys are awesome.
5C3 is only 10 so that doesn't seem to be correct though this may be cause of the weird fact that 5C3 and 5C2 are both 10... maybe for 6 points and greater the formula works... although for 5 points and below it would be quite easy to figure it out without using a formula... anyone with a clearer idea?
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