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@Anurag is this property also applicable to any polygon with n sides?

I used same property and concluded that PT < 14 but using other property
The length of each side is greater than the difference between other 2 sides i found 10 < PT . What is INCORRECT here as it does not satisfy any option

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yep refer to http://www.beatthegmat.com/a-quickier-way-t87336.html#463160 for a discussion on the lower limit of last side

You can not follow that method because:

1) You might disregard the important condition of the polygon being complex if you blindly add the limits of the diagonals by apply the triangle rule to the inner triangles.
2) You assume that the minimum of the last side will happen when the diagonals are at minimum length too. That might not be the case
For example, in the polygon described in your question, there is no lower limit for the last side. PT can be tending to 0 and PQRS could be a valid 4 sided Convex Polygon- A quadrilateral.

Triangle rule extended to pentagons

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Thanks Mr Smith but unfortunately i am not able to comprehend both of the responses. Please explain in simpler language.
My concern is how would one solve this problem on GMAT DAY Even OG explanation assumes lot of stuff, it seems that they trying to prove ans and not derive it. for proving 5 to be a valid case they take different numbers and for 10 they chose different approach

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Ok heres How I would approach the problem on GMAT Day.

given 4 side lengths: 5,4,3,2
Thus maximum length of fifth side: 14
Thus 15 is out

Now to find if there is minimum constraint on the 5th side, As GMATguruNY aptly pointed out, Lets find out if the remaining sides can make quadrilateral. Find out if the largest side length is greater than the sum of the remaining 3 side lengths. If yes then find the difference which is the minimum value of the 5th side. Since, in that case, the remaining 4 sides cannot form a valid quadrilateral, the fifth side is restricted by minimum limit for a valid convex pentagon.

In our Case, 5<(4+3+2) thus the four sides can form a valid quadrilateral PQRS thus PT can even tend to zero for a valid pentagon
So 0<PT<14.
Thus 5 and 10 is the answer

Thanks Mitch and Mr.Smith

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