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

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

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New post 12 Sep 2016, 09:56
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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


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Re: An equilateral triangle T2 is formed by joining the mid points of the  [#permalink]

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New post 12 Sep 2016, 10:29
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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.
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Re: An equilateral triangle T2 is formed by joining the mid points of the  [#permalink]

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New post 08 Feb 2018, 04:49
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...."
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Re: An equilateral triangle T2 is formed by joining the mid points of the  [#permalink]

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New post 08 Feb 2018, 06:00
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
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Re: An equilateral triangle T2 is formed by joining the mid points of the  [#permalink]

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New post 08 Feb 2018, 08:41
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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 !
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Re: An equilateral triangle T2 is formed by joining the mid points of the &nbs [#permalink] 08 Feb 2018, 08:41
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