Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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.
_________________
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.
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.
ABCDE is a regular pentagon with F at its center. How many [#permalink]
Show Tags
28 May 2012, 01:03
2
6
macjas 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
Generally in a plane if there are \(n\) points of which no three are collinear, then: 1. The number of triangles that can be formed by joining them is \(C^3_n\).
2. The number of quadrilaterals that can be formed by joining them is \(C^4_n\).
3. The number of polygons with \(k\) sides that can be formed by joining them is \(C^k_n\).
Since ABCDE is a regular pentagon then no three point out of 6 (5 vertices + center) will be collinear, so the number of triangles possible is \(C^3_6=20\).
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]
Show Tags
25 Dec 2013, 03:15
1
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
Re: ABCDE is a regular pentagon with F at its center. How many d [#permalink]
Show Tags
16 Feb 2018, 07:04
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
close
60% (00:44) correct 
