Pentagon sides

[vikram4689]

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 10only
D 10 and 15 only
E 5,10 and 15
OA C

RULE [/b]The length of each side is
less than the sum of the lengths of the other two sides and
greater than the difference between these lengths

Join PR and PS. Use above property to find
1 < PR < 5
5 < PS < 9 and hence
10 < PT < 14

What is wrong in above method

[Anurag@Gurome]

We know that the length of any side of a triangle should always be smaller than the sum of other two sides.

Similarly, the length of any side of a pentagon should be smaller than the sum of other 4 sides.
So, PQ + QR + RS + ST = 3 + 2 + 4 + 5 = 14, which means the length of 5th side should not be more than 14.

The correct answer is C.

[shubhamkumar]

We know that the length of any side of a triangle should always be smaller than the sum of other two sides.

Similarly, the length of any side of a pentagon should be smaller than the sum of other 4 sides.
So, PQ + QR + RS + ST = 3 + 2 + 4 + 5 = 14, which means the length of 5th side should not be more than 14.

The correct answer is C.

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

[GMATGuruNY]

We know that the length of any side of a triangle should always be smaller than the sum of other two sides.

Similarly, the length of any side of a pentagon should be smaller than the sum of other 4 sides.
So, PQ + QR + RS + ST = 3 + 2 + 4 + 5 = 14, which means the length of 5th side should not be more than 14.

The correct answer is C.

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

Yes. The length of any side of a polygon with n sides must be less than the sum of the lengths of the other sides.

I posted an explanation for this problem here:

https://www.beatthegmat.com/a-quickier-w ... tml#463160

[vikram4689]

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

[Mr Smith]

yep refer to https://www.beatthegmat.com/a-quickier-w ... tml#463160 for a discussion on the lower limit of last side

[Mr Smith]

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.

[ronnie1985]

Triangle rule extended to pentagons

[vikram4689]

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

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.

[GMATGuruNY]

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

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.

Use common sense.

It is possible to form a quadrilateral with sides of 3, 2, 4 and 5, since the length of each side would be less than the sum of the lengths of the other three sides.

Thus, it must be possible to add a 5th side that is INFINITELY SMALL:
1. Select any vertex.
2. Separate -- just a little -- the two sides forming that vertex.
3. Connect the separated sides with a tiny 5th side.

Click on the link in my post above for an illustration of just such a polygon.

[Mr Smith]

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

[vikram4689]

Thanks Mitch and Mr.Smith