Hey Everybody,
In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the lenghts 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
the explanation given by the book is very logic but takes a very long time. Because they told us to try each of the explanation possible and each time we have to make assumptions for example by changing the figures.
So my question is " Is there a quickier way to solve it without spend 3 or 4 minutes on it ? "
Thanks
A quickier way ?
This topic has expert replies
Solution:5 and 10 only
This is how the book make its reasoning:'first cut the pentagone in 3 triangle making a connection between QT and RT. So we have three triangle QPT, QRT and RTS. We know also that the lenght of any side is less than the sum of the lenght of the other two side
Ametting 5 is a solution it said that QT can be equal to 7 because with QP=3 and PT=5 the inequality is verified
QT=7< 8= 5+3= QP + PT
PQ=3< 12 =7+5= PT+TQ
PT=5<10 = 3+7= PQ+ TQ
For the triangle QRT the same technique is applied since QT is supposed to be 7 we have
RT= 8<9= 7+2 =QT+QR
RQ= 2< 15 = 7+8=QT+RT
QT=7< 10= 2+8 = QR+RT
so 5 is a solution because it checked the inequality
Then repeat the same process for the 10 and 15"
Source Quantitative review 2nd edition
This is how the book make its reasoning:'first cut the pentagone in 3 triangle making a connection between QT and RT. So we have three triangle QPT, QRT and RTS. We know also that the lenght of any side is less than the sum of the lenght of the other two side
Ametting 5 is a solution it said that QT can be equal to 7 because with QP=3 and PT=5 the inequality is verified
QT=7< 8= 5+3= QP + PT
PQ=3< 12 =7+5= PT+TQ
PT=5<10 = 3+7= PQ+ TQ
For the triangle QRT the same technique is applied since QT is supposed to be 7 we have
RT= 8<9= 7+2 =QT+QR
RQ= 2< 15 = 7+8=QT+RT
QT=7< 10= 2+8 = QR+RT
so 5 is a solution because it checked the inequality
Then repeat the same process for the 10 and 15"
Source Quantitative review 2nd edition
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The thirdside rule of triangles states that the third side of any triangle must be less than the sum of the lengths of the other 2 sides.obokan wrote:Hey Everybody,
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
the explanation given by the book is very logic but takes a very long time. Because they told us to try each of the explanation possible and each time we have to make assumptions for example by changing the figures.
So my question is " Is there a quickier way to solve it without spend 3 or 4 minutes on it ? "
Thanks
This logic extends to any polygon:
The last side of any polygon must be less than the sum of the lengths of the other sides.
Given sides of 3,2,4 and 5, the length of the 5th side must be less than 3+2+4+5 = 14.
Eliminate D and E.
No reason the 5th side couldn't be 5 or 10.
The correct answer is C.
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Thanks Mitch, Didn't think of it this way!GMATGuruNY wrote:
The thirdside rule of triangles states that the third side of any triangle must be less than the sum of the lengths of the other 2 sides.
This logic extends to any polygon:
The last side of any polygon must be less than the sum of the lengths of the other sides.

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Is there a simple/intuitive way to say 5 is definitely an answer, as there has to be some kind of lower limit to the length of PT.
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ahsgirlie87 wrote:Is there a simple/intuitive way to say 5 is definitely an answer, as there has to be some kind of lower limit to the length of PT.
The figure above illustrates that PT can be infinitely small.
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very good ahsgirlie87, there certainly is a lower limit
Yes the methods described above are certainly correct and work out very well for finding the upper limit of unknown side. I feel that questions asking or testing the lower limit of the side for a polygon would become too complex and would be beyond the scope of GMAT
Having said that, for all those who are curious.
There will be a minimum length for Nth side of the N sided polygon if for any side of length x, the sum of the lengths of remaining N2 sides S is less than x Basically if there are two or more acute angle in the polygon. And the minimum length shall be the difference between x and S
Note: This will happen to maximum one side
How to check this: Check if the length of the largest given side is greater than the sum of lengths of remaining N2 given sides. If it is, find the difference. The minimum length of the unknown side will be this difference
This is derived basically from the concepts of:
3rd side of a triangle is always between the sum and the difference of sides.
In our example, the largest given length is 5 which is not greater than the sum of other given lengths 5<(4+3+2)
Hence in our case the last side has no minimum.
If for example the length of ST = 10, then the minimum length of PT would be 10(4+3+2)=1, then 1<PT<14
Yes the methods described above are certainly correct and work out very well for finding the upper limit of unknown side. I feel that questions asking or testing the lower limit of the side for a polygon would become too complex and would be beyond the scope of GMAT
Having said that, for all those who are curious.
There will be a minimum length for Nth side of the N sided polygon if for any side of length x, the sum of the lengths of remaining N2 sides S is less than x Basically if there are two or more acute angle in the polygon. And the minimum length shall be the difference between x and S
Note: This will happen to maximum one side
How to check this: Check if the length of the largest given side is greater than the sum of lengths of remaining N2 given sides. If it is, find the difference. The minimum length of the unknown side will be this difference
This is derived basically from the concepts of:
3rd side of a triangle is always between the sum and the difference of sides.
In our example, the largest given length is 5 which is not greater than the sum of other given lengths 5<(4+3+2)
Hence in our case the last side has no minimum.
If for example the length of ST = 10, then the minimum length of PT would be 10(4+3+2)=1, then 1<PT<14