GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Oct 2019, 14:20

### 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

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

# An equilateral triangle T2 is formed by joining the mid points of the

Author Message
TAGS:

### Hide Tags

Manager
Joined: 17 Jun 2015
Posts: 195
GMAT 1: 540 Q39 V26
GMAT 2: 680 Q46 V37
An equilateral triangle T2 is formed by joining the mid points of the  [#permalink]

### Show Tags

12 Sep 2016, 10:56
3
6
00:00

Difficulty:

45% (medium)

Question Stats:

69% (02:18) correct 31% (02:27) wrong based on 109 sessions

### HideShow timer Statistics

An equilateral triangle T2 is formed by joining the mid points of the sides of another equilateral triangle T1. A third equilateral triangle T3 is formed by joining the mid-points of T2 and this process is continued indefinitely. If each side of T1 is 40 cm, find the sum of the perimeters of all the triangles.

A. 180 cm
B. 220 cm
C. 240 cm
D. 270 cm
E. 300 cm

In dire need of kudos.

_________________
Fais de ta vie un rêve et d'un rêve une réalité
Retired Moderator
Joined: 26 Nov 2012
Posts: 569
Re: An equilateral triangle T2 is formed by joining the mid points of the  [#permalink]

### Show Tags

12 Sep 2016, 11:29
4
2
hdwnkr wrote:
An equilateral triangle T2 is formed by joining the mid points of the sides of another equilateral triangle T1. A third equilateral triangle T3 is formed by joining the mid-points of T2 and this process is continued indefinitely. If each side of T1 is 40 cm, find the sum of the perimeters of all the triangles.

A. 180 cm
B. 220 cm
C. 240 cm
D. 270 cm
E. 300 cm

In dire need of kudos.

We have 40 for first triangle, when we join mid-points of first triangle we get the second equilateral traingle then the length of second one is 20 and continues.

So we have 40,20,10,...

We have ratio = 1/2, and it is GP type.

Sum of infinite triangle is a/1-r = 40/1-(1/2) = 80

Equilateral triangle perimeter is 3a = 3*80 = 240.

So option C.

Let me know if I am missing anything.
##### General Discussion
Director
Joined: 27 May 2012
Posts: 902
Re: An equilateral triangle T2 is formed by joining the mid points of the  [#permalink]

### Show Tags

08 Feb 2018, 05:49
msk0657 wrote:
hdwnkr wrote:
An equilateral triangle T2 is formed by joining the mid points of the sides of another equilateral triangle T1. A third equilateral triangle T3 is formed by joining the mid-points of T2 and this process is continued indefinitely. If each side of T1 is 40 cm, find the sum of the perimeters of all the triangles.

A. 180 cm
B. 220 cm
C. 240 cm
D. 270 cm
E. 300 cm

In dire need of kudos.

We have 40 for first triangle, when we join mid-points of first triangle we get the second equilateral traingle then the length of second one is 20 and continues.

So we have 40,20,10,...

We have ratio = 1/2, and it is GP type.

Sum of infinite triangle is a/1-r = 40/1-(1/2) = 80

Equilateral triangle perimeter is 3a = 3*80 = 240.

So option C.

Let me know if I am missing anything.

Dear Karishma,chetan2u,

Can any of you explain this in a better way, not able to comprehend " Sum of an infinite triangle ..is a/1-r...."
_________________
- Stne
Manager
Joined: 28 Jan 2018
Posts: 50
Location: Netherlands
Concentration: Finance
GMAT 1: 710 Q50 V36
GPA: 3
Re: An equilateral triangle T2 is formed by joining the mid points of the  [#permalink]

### Show Tags

08 Feb 2018, 07:00
I did this problem with approximation.

Each subsequent triangle will have the side equal to 1/2 of the previous triangle, hence the sum of their perimeter will be:

(40+20+10+5+2.5+1.25+.....) * 3
observing the pattern we can see that the perimeter of each triangle after .... will be < 1 and become closer and closer to 0, therefore the answer must be 240 (80 * 3), but i don't know how to solve this otherwise
Director
Joined: 27 May 2012
Posts: 902
Re: An equilateral triangle T2 is formed by joining the mid points of the  [#permalink]

### Show Tags

08 Feb 2018, 09:41
1
hoang221 wrote:
I did this problem with approximation.

Each subsequent triangle will have the side equal to 1/2 of the previous triangle, hence the sum of their perimeter will be:

(40+20+10+5+2.5+1.25+.....) * 3
observing the pattern we can see that the perimeter of each triangle after .... will be < 1 and become closer and closer to 0, therefore the answer must be 240 (80 * 3), but i don't know how to solve this otherwise

Dear hoang221,

The sum of an infinite converging geometric series = $$\frac{ a}{1-r}$$ where a is the first term and r is the common ratio.

Since the triangles keep converging infinitely we can use this formula.
check this link for more on this:

http://www.nabla.hr/IA-SequenceAndSerie3.htm

Hope this helps !
_________________
- Stne
Non-Human User
Joined: 09 Sep 2013
Posts: 13260
Re: An equilateral triangle T2 is formed by joining the mid points of the  [#permalink]

### Show Tags

30 Mar 2019, 08:08
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.
_________________
Re: An equilateral triangle T2 is formed by joining the mid points of the   [#permalink] 30 Mar 2019, 08:08
Display posts from previous: Sort by