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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?

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

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20 Nov 2016, 08:08

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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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29 Aug 2017, 09:06

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!