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Re: Pentagon problem [#permalink]
03 Nov 2009, 15:10

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Expert's post

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arora2m 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?

a. 10

b. 15

c. 20

d. 25

e. 30

Any 3 points from 6 will make a triangle, since no 3 points are collinear, then:

6C3=20

Answer: C.

TIPS ON RELATED ISSUES: In a plane if there are n points of which no three are collinear, then 1. The number of straight lines that can be formed by joining them is nC2. 2. The number of triangles that can be formed by joining them is nC3. 3. The number of polygons with k sides that can be formed by joining them is nCk. _________________

Re: Pentagon problem [#permalink]
03 Nov 2009, 15:39

Bunuel - In this case I am a bit confused. F is the center of the pentagon ABCDE ... So ABF form a triangle but not ACF, since a straight line drawn between AC will not pass through F.

I got 15 by doing 5 triangles formed with F as one of the points and then 5C3 thus 5+10 =15.

Please tell me what is wrong with my reasoning here.

Re: Pentagon problem [#permalink]
03 Nov 2009, 15:55

arora2m wrote:

Bunuel - In this case I am a bit confused. F is the center of the pentagon ABCDE ... So ABF form a triangle but not ACF, since a straight line drawn between AC will not pass through F.

I got 15 by doing 5 triangles formed with F as one of the points and then 5C3 thus 5+10 =15.

Please tell me what is wrong with my reasoning here.

acf does form a triangle. think of F immediately under A so drop a vertical line from a to f and then draw a line form a to c then connect c to f

the line from a to f stops at f and doesn't continue on to another letter. it angles over to c

Re: Pentagon problem [#permalink]
03 Nov 2009, 16:01

Expert's post

arora2m wrote:

Bunuel - In this case I am a bit confused. F is the center of the pentagon ABCDE ... So ABF form a triangle but not ACF, since a straight line drawn between AC will not pass through F.

I got 15 by doing 5 triangles formed with F as one of the points and then 5C3 thus 5+10 =15.

Please tell me what is wrong with my reasoning here.

Regular pentagon is a pentagon where all sides are equal. In such pentagon center is not collinear to any two vertex, so ANY three points (from 5 vertices and center point) WILL form the triangle.

You wrote that ACF won't form the triangle because they don't lie on straight line - that's not true. EXACTLY because these three points DON'T lie on the straight line they WILL form triangle. (In your other example you stated that ABF will form the triangle, but these three points also aren't collinear).

The question basically asks how many triangles can be formed from the six points on a plane with no three points being collinear.

Re: ABCDE is a regular pentagon with F at its center. How many d [#permalink]
15 Aug 2013, 06:03

arora2m 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?

A. 10 B. 15 C. 20 D. 25 E. 30

ITS 6C3 = 20 . AND THERE ARE NO POINTS THOSE WILL NOT FORM TRIANGLE................... ABC, BCD ALL THESE 5 DIFFERENT COMBINATIONS OF POINTS WILL FORM TRIANGLE... SO ANSWER IS 20 _________________

Re: Pentagon problem [#permalink]
21 Aug 2013, 11:29

Bunuel wrote:

arora2m 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?

a. 10

b. 15

c. 20

d. 25

e. 30

Any 3 points from 6 will make a triangle, since no 3 points are collinear, then:

6C3=20

Answer: C.

TIPS ON RELATED ISSUES: In a plane if there are n points of which no three are collinear, then 1. The number of straight lines that can be formed by joining them is nC2. 2. The number of triangles that can be formed by joining them is nC3. 3. The number of polygons with k sides that can be formed by joining them is nCk.

Re: Pentagon problem [#permalink]
21 Aug 2013, 11:31

Expert's post

honchos wrote:

Bunuel wrote:

arora2m 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?

a. 10

b. 15

c. 20

d. 25

e. 30

Any 3 points from 6 will make a triangle, since no 3 points are collinear, then:

6C3=20

Answer: C.

TIPS ON RELATED ISSUES: In a plane if there are n points of which no three are collinear, then 1. The number of straight lines that can be formed by joining them is nC2. 2. The number of triangles that can be formed by joining them is nC3. 3. The number of polygons with k sides that can be formed by joining them is nCk.

Where am I wrong, I used, 5C2 X 1 = 10

You should explain the logic behind your approach. _________________

Re: Pentagon problem [#permalink]
21 Aug 2013, 11:48

1

This post received KUDOS

Expert's post

honchos wrote:

ABCDE, 5 vertices of pentagon, we can select any two vertices two form a line, 5C2 ways....

Now any of the 5C2 ways have only one way(As per question) to interact with a centre point F = 5C2 X 1 = 10 Ways.

How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?

So, we are interested in the number of triangles that can be formed by joining ANY three out of these 6 points (A, B, C, D, E and F), not necessarily F and 2 other points. For example, triangle ABC is also a valid one.

Re: ABCDE is a regular pentagon with F at its center. How many d [#permalink]
25 Dec 2013, 02:15

1

This post received KUDOS

This is basically a combinatorics problem. If we make anagram grid for this, it will like below A B C D E F Y Y Y N N N That means at any point of time only three points are making the triangle while other three are not. hence total no of combinations possible = 6!/3!x3! =20

gmatclubot

Re: ABCDE is a regular pentagon with F at its center. How many d
[#permalink]
25 Dec 2013, 02:15

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