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by pradeepkaushal9518 » Sun Jul 04, 2010 4:53 am
ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?



10



15



20



25



30
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by kvcpk » Sun Jul 04, 2010 5:51 am
pradeepkaushal9518 wrote:ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?
Is the OA 6c3 = 20??
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by pradeepkaushal9518 » Sun Jul 04, 2010 7:18 am
no kvcpk 20 is not the answer.i dont know the method to solve it
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by shankysainik » Sun Jul 04, 2010 7:23 am
Is it 15?

5C3 = 10 (For any 3 points among A,B,C,D,E)
and 5 for the number of triangles possible using the centre F)

Please let me know if this is right!
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by kvcpk » Sun Jul 04, 2010 7:32 am
pradeepkaushal9518 wrote:no kvcpk 20 is not the answer.i dont know the method to solve it
I still believe answer should be only 6c3 = 20. Because, any 3 points can form the triangle..

What is the OA?
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by kvcpk » Sun Jul 04, 2010 7:35 am
shankysainik wrote:Is it 15?

5C3 = 10 (For any 3 points among A,B,C,D,E)
and 5 for the number of triangles possible using the centre F)

Please let me know if this is right!
Hi,

ABCDE is a regular pentagon.. So F will not fall in line with any lines drawn between two sides. Hence F can form a triangle with any of the two vertices...

You missed out the triangles ACF, ADF,etc..

What you say??
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by mj78ind » Sun Jul 04, 2010 7:39 am
kvcpk wrote:
shankysainik wrote:Is it 15?

5C3 = 10 (For any 3 points among A,B,C,D,E)
and 5 for the number of triangles possible using the centre F)

Please let me know if this is right!
Hi,

ABCDE is a regular pentagon.. So F will not fall in line with any lines drawn between two sides. Hence F can form a triangle with any of the two vertices...

You missed out the triangles ACF, ADF,etc..

What you say??
I am with kvcpk, we have 6 vertices with no three vertices in the same line hence we can have 6C3 triangles.......
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by shankysainik » Sun Jul 04, 2010 7:53 am
Agreed- should be 6C3- my bad!
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by pradeepkaushal9518 » Sun Jul 04, 2010 9:08 am
sorry kvcpk answer is 20 only i have choosen 15 which was wrong.
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by kvcpk » Sun Jul 04, 2010 9:54 am
pradeepkaushal9518 wrote:sorry kvcpk answer is 20 only i have choosen 15 which was wrong.
No Problem... Happy to see that what I thought was right ;)
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