Last visit was: 25 Apr 2024, 07:31 It is currently 25 Apr 2024, 07:31

Close
GMAT Club Daily Prep
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.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Kudos
Tags:
Show Tags
Hide Tags
avatar
Director
Director
Joined: 29 Nov 2012
Posts: 580
Own Kudos [?]: 6042 [20]
Given Kudos: 543
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 92914
Own Kudos [?]: 618951 [16]
Given Kudos: 81595
Send PM
User avatar
Director
Director
Joined: 02 Sep 2012
Status:Far, far away!
Posts: 859
Own Kudos [?]: 4891 [10]
Given Kudos: 219
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Send PM
General Discussion
User avatar
Manager
Manager
Joined: 25 Feb 2013
Status:*Lost and found*
Posts: 102
Own Kudos [?]: 196 [0]
Given Kudos: 14
Location: India
Concentration: General Management, Technology
GMAT 1: 640 Q42 V37
GPA: 3.5
WE:Web Development (Computer Software)
Send PM
Re: How many triangles can be inscribed in the heptagon pictured [#permalink]
fozzzy wrote:
Attachment:
heptagon.png
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

A. 9
B. 21
C. 35
D. 140
E. 210


I guess another way to look at it is that if we are to consider any line under the heptagon, we can see there are 5 other vertices left with which the line can form a traingle. For example for a line composed of AB, there are vertices C,D,E,F,G with which traingles can be constructed.

In total there are 7 possible lines fo the heptagon. Hence total traingles = 7*5 = 35. Hence [C].

Regards,
A
avatar
Director
Director
Joined: 29 Nov 2012
Posts: 580
Own Kudos [?]: 6042 [0]
Given Kudos: 543
Send PM
Re: How many triangles can be inscribed in the heptagon pictured [#permalink]
so in the same question if we were asked how many quadrilaterals to be formed it would be 7C4?
Math Expert
Joined: 02 Sep 2009
Posts: 92914
Own Kudos [?]: 618951 [0]
Given Kudos: 81595
Send PM
Re: How many triangles can be inscribed in the heptagon pictured [#permalink]
Expert Reply
fozzzy wrote:
so in the same question if we were asked how many quadrilaterals to be formed it would be 7C4?


Yes, that's correct.
Manager
Manager
Joined: 08 Feb 2016
Posts: 53
Own Kudos [?]: 60 [0]
Given Kudos: 25
Location: India
Concentration: Technology
Schools: AGSM '20 (A)
GMAT 1: 650 Q49 V30
GPA: 4
Send PM
Re: How many triangles can be inscribed in the heptagon pictured [#permalink]
Bunuel,

I don't seem to remember formulas during exam. Can you please point out the flaw in my approach ?

We are required to make triangles inside the heptagon such that three vertices of the triangle are also vertices of the heptagon.
This means we are simply drawing lines to connect the vertices of the heptagon, connecting 2 vertices of the heptagon at a time (that's how we form the triangles as directed in the problem)

That means we have to choose 2 points at a time from 7 points. That is \(C^2_7=21\).

Bunuel wrote:
fozzzy wrote:
Attachment:
heptagon.png
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

A. 9
B. 21
C. 35
D. 140
E. 210


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 no 3 vertices in given heptagon are collinear, then the number of triangles possible is \(C^3_7=35\).

Answer: C.

Similar questions to practice:
https://gmatclub.com/forum/if-4-points-a ... 32677.html
https://gmatclub.com/forum/abcde-is-a-re ... 86284.html
https://gmatclub.com/forum/abcde-is-a-re ... 33328.html
https://gmatclub.com/forum/the-sides-bc- ... 09690.html
https://gmatclub.com/forum/gmat-diagnost ... 79373.html
https://gmatclub.com/forum/m03-71107.html
https://gmatclub.com/forum/how-many-tria ... 98236.html
https://gmatclub.com/forum/right-triangl ... 88958.html
https://gmatclub.com/forum/how-many-circ ... 28149.html

Hope it helps.
Math Expert
Joined: 02 Sep 2009
Posts: 92914
Own Kudos [?]: 618951 [0]
Given Kudos: 81595
Send PM
Re: How many triangles can be inscribed in the heptagon pictured [#permalink]
Expert Reply
ajay2121988 wrote:
Bunuel,

I don't seem to remember formulas during exam. Can you please point out the flaw in my approach ?

We are required to make triangles inside the heptagon such that three vertices of the triangle are also vertices of the heptagon.
This means we are simply drawing lines to connect the vertices of the heptagon, connecting 2 vertices of the heptagon at a time (that's how we form the triangles as directed in the problem)

That means we have to choose 2 points at a time from 7 points. That is \(C^2_7=21\).

Bunuel wrote:
fozzzy wrote:
Attachment:
heptagon.png
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

A. 9
B. 21
C. 35
D. 140
E. 210


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 no 3 vertices in given heptagon are collinear, then the number of triangles possible is \(C^3_7=35\).

Answer: C.

Similar questions to practice:
https://gmatclub.com/forum/if-4-points-a ... 32677.html
https://gmatclub.com/forum/abcde-is-a-re ... 86284.html
https://gmatclub.com/forum/abcde-is-a-re ... 33328.html
https://gmatclub.com/forum/the-sides-bc- ... 09690.html
https://gmatclub.com/forum/gmat-diagnost ... 79373.html
https://gmatclub.com/forum/m03-71107.html
https://gmatclub.com/forum/how-many-tria ... 98236.html
https://gmatclub.com/forum/right-triangl ... 88958.html
https://gmatclub.com/forum/how-many-circ ... 28149.html

Hope it helps.


Sorry don't follow you. We need three vertices for a triangle not two.
Manager
Manager
Joined: 08 Feb 2016
Posts: 53
Own Kudos [?]: 60 [0]
Given Kudos: 25
Location: India
Concentration: Technology
Schools: AGSM '20 (A)
GMAT 1: 650 Q49 V30
GPA: 4
Send PM
Re: How many triangles can be inscribed in the heptagon pictured [#permalink]
It's alright, I was confusing this with another question. All clear now. Thanks.

Bunuel wrote:
ajay2121988 wrote:
Bunuel,

I don't seem to remember formulas during exam. Can you please point out the flaw in my approach ?

We are required to make triangles inside the heptagon such that three vertices of the triangle are also vertices of the heptagon.
This means we are simply drawing lines to connect the vertices of the heptagon, connecting 2 vertices of the heptagon at a time (that's how we form the triangles as directed in the problem)

That means we have to choose 2 points at a time from 7 points. That is \(C^2_7=21\).



Sorry don't follow you. We need three vertices for a triangle not two.
Current Student
Joined: 23 Apr 2018
Posts: 130
Own Kudos [?]: 63 [0]
Given Kudos: 176
Send PM
Re: How many triangles can be inscribed in the heptagon pictured [#permalink]
Bunuel wrote:
fozzzy wrote:
so in the same question if we were asked how many quadrilaterals to be formed it would be 7C4?


Yes, that's correct.



and that will also give the same answer as number of triangles.. as 35 ?
7C4
Intern
Intern
Joined: 28 Jun 2016
Posts: 1
Own Kudos [?]: 5 [0]
Given Kudos: 1
Location: India
Concentration: Operations, Strategy
Schools: Fisher '17
GMAT 1: 710 Q49 V38
GPA: 3.4
WE:Operations (Transportation)
Send PM
Re: How many triangles can be inscribed in the heptagon pictured [#permalink]
things become interesting when specific number of sides of heptagon is asked 2 include
Director
Director
Joined: 09 Mar 2018
Posts: 783
Own Kudos [?]: 453 [0]
Given Kudos: 123
Location: India
Send PM
Re: How many triangles can be inscribed in the heptagon pictured [#permalink]
fozzzy wrote:
Attachment:
heptagon.png
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

A. 9
B. 21
C. 35
D. 140
E. 210


Since you need 3 non collinear points to make a triangle

7C3 = 35

Answer C
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18758
Own Kudos [?]: 22051 [0]
Given Kudos: 283
Location: United States (CA)
Send PM
Re: How many triangles can be inscribed in the heptagon pictured [#permalink]
Expert Reply
fozzzy wrote:
Attachment:
heptagon.png
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

A. 9
B. 21
C. 35
D. 140
E. 210


Since there are 7 vertices on the heptagon and any of its 3 vertices will form a triangle, then the number of possible triangles is 7C3 = 7!/(3! x 4!) = (7 x 6 x 5)/3! = (7 x 6 x 5)/(3 x 2) = 35.

Answer: C
Intern
Intern
Joined: 09 Apr 2017
Posts: 41
Own Kudos [?]: 17 [0]
Given Kudos: 42
Location: India
Send PM
Re: How many triangles can be inscribed in the heptagon pictured [#permalink]
what about the repeating Triangles? Should we not take that into consideration?
Side GA can make 5 triangles, and seven sides can produce 7*5= 35, but GAB and ABG are the same triangles.

please advise on this.
GMAT Club Bot
Re: How many triangles can be inscribed in the heptagon pictured [#permalink]
Moderators:
Math Expert
92914 posts
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne