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In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the

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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]

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New post 09 Nov 2015, 17:19
russ9 wrote:
Bunuel wrote:
MensaNumber wrote:
Hi Experts,

Yes please could you tell us if there is a rule for the minimum possible length in a polygon (like the one we have for triangle)?

Thanks!


Please check here: in-the-pentagon-pqrst-which-is-attached-as-a-jpg-file-pq-126943.html#p1051688 an here: in-the-pentagon-pqrst-which-is-attached-as-a-jpg-file-pq-126943.html#p1214245

Hope this helps.


Hi Bunuel,

To piggy back on this question -- the links that you have provided state that the length of the 5th side has to be less than the sum of the other 4 sides. Is there anything governing the lower end? For example, in a triangle, the third side must be bigger than the difference of the two sides. Does a similar rule apply to pentagons and if they do, how do you compare which 2 sides?

My question just lies as to why we didn't prove/justify that 5 wasn't too small? In this case, we don't have an answer choice that just states that "10" is the value but if it did, are we to assume that 5 will automatically work?


Hi Bunuel, I have the same question as russ9. Is there a minimum length for the side of a polygon?

Thanks so much!

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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]

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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]

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New post 13 May 2017, 07:55
victory47 wrote:
I do not see this question from og books.

pls tell me the number of this question.

OG question is official guide.



No. 149 Official Guide Math Review 2017
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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]

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New post 03 Jun 2017, 01:15
abhi47 wrote:
Attachment:
Pentagon PQRST.PNG
In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the lengths 5, 10 and 15 could be the value of PT?

A 5 only
B 10 only
C 5 and 10 only
D 10 and 15 only
E 5, 10 and 15

OPEN DISCUSSION OF THIS QUESTION IS HERE: http://gmatclub.com/forum/in-pentagon-p ... 68634.html


hey,

is it also true that PT must be > 4 ?
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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]

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New post 29 Aug 2017, 10:06
Bunuel wrote:
pbull78 wrote:
bunuel any other wayout for this problem. i have drawn 3 triangles and done that way, is it ok. thanks


Yes, you can solve it with a triangle property which says that the length of any side of a triangle must be smaller than the sum of the other two sides.

Though you can naturally apply this property to any polygon and you won't need intermediary steps for solving.


This is easier to understand than in explanation in OG.
Thanks a lot!

Posted from my mobile device

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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the   [#permalink] 29 Aug 2017, 10:06

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