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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]
29 Feb 2012, 06:50

1

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Expert's post

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. _________________

Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]
08 Sep 2012, 02:29

As mentioned by Bunuel, I will give another way to solve this problem using Triangle property.

Join P & R In triangle PQR, the known two sides are 3 & 2 units so third side (Let x) must be between 1<x<5 i.e. it can take any value 2,3 or 4 In other words the maximum & minimum values possible are 4 & 2 respectively

Join T & R In triangle PQR, the known two sides are 4 & 5 units so third side (Let y) must be between 1<x<9 i.e. it can take any value 2,3,4,5,6,7 or 8 In other words the maximum & minimum values possible are 8 & 2 respectively

We want to know the values possible for PT In triangle PRT, the range of two sides is 1<x<5 & 1<x<9 units so third side (PT) must be between (2-2)< PT <(8+4) or 0< PT<12 i.e. it can take any value 1,2,3,4,5,6,7....10,11 (assuming sides can take only integer values)

Thus both 5 & 10 are possible Thus Answer C

Hope it helps. _________________

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No posting of PS/DS questions is allowed in the main Math forum.

Hi Bunuel,

The length of any side of a triangle must smaller than the sum of the other two sides. The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides.

is this true only for triangle and pentagon or any other polygon? How can we determine it? _________________

GMAT - Practice, Patience, Persistence Kudos if u like

Re: In the pentagon PQRST which is attached as a jpg file , PQ=3 [#permalink]
18 Apr 2013, 06:50

2

This post received KUDOS

Bunuel wrote:

The length of any side of a triangle must smaller than the sum of the other two sides. The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides.

Dear Bunuel, does this property apply to all quadrilaterals ? -- Thank you _________________

Re: In the pentagon PQRST which is attached as a jpg file , PQ=3 [#permalink]
19 Apr 2013, 02:55

1

This post received KUDOS

Expert's post

jainpiyushjain wrote:

Bunuel wrote:

The length of any side of a triangle must smaller than the sum of the other two sides. The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides.

Dear Bunuel, does this property apply to all quadrilaterals ? -- Thank you

Yes, the length of any side of a polygon must be less than the sum of the lengths of the other sides. _________________

Re: In Pentagon PQRST, PQ = 3,OR =5,RS =2 and ST =5 [#permalink]
17 May 2013, 06:32

1

This post received KUDOS

SrinathVangala wrote:

In Pentagon PQRST, PQ = 3,OR =5,RS =2 and ST =5.Which of the lengths 5,10 and 15 could be the value of PT?

Attachment:

The attachment Untitled.jpg is no longer available

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

Good one! The answer seems to be [C].

The easy way to approach this would be to consider the pentagon be comprised of 3 triangles as in the figure.

Let x and y be the diagonals under the quad which make the 3 triangles possible. Now by triangle property, we know x < QR + PQ = 8 ......(1) and y < RS + ST = 7 ...........(2)

Hence now in the interior triangle with sides x,y and PT, we further apply the triangle property: PT < x+y From above we already know by adding (1) and (2) x+y < 15, Hence we can say, that PT < 15. That rules out 15 off the options and 5, 10 can be possible values of the side!

Re: In Pentagon PQRST, PQ = 3,OR =5,RS =2 and ST =5 [#permalink]
17 May 2013, 06:36

Since there is no more information is given about the pentagon, only condition which has to be satisfied is that sum of any 4 sides must be larger than the fifth side. Now, 4 lengths are given as: 2,3,5,5. Fifth side must be smaller than sum of these 4, i.e must be smaller than 15, so 15 can't be the answer. About 5 and 10, there can not be any restriction, as the above rule is not broken in any of these cases. So, answer is c) 5 and 10 only.

Re: In Pentagon PQRST, PQ = 3,OR =5,RS =2 and ST =5 [#permalink]
17 May 2013, 06:40

Expert's post

SrinathVangala wrote:

In Pentagon PQRST, PQ = 3,OR =5,RS =2 and ST =5.Which of the lengths 5,10 and 15 could be the value of PT?

Attachment:

Untitled.jpg

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

Assuming OR is actually QR above, then this problem hinges on the fact that the fifth side of the pentagon must be shorter than the sum of the other four sides. If the four sides add up to 15 (3+5+2+5) then the fifth side cannot be 15, as this would form a line. Similarly, it cannot be greater than 15. It could, however, easily be 5 or 10, depending on the angles of the other arcs. Answer choice C.

Re: In the pentagon PQRST which is attached as a jpg file , PQ=3 [#permalink]
16 Jan 2014, 02:33

1

This post received KUDOS

Bunuel wrote:

jainpiyushjain wrote:

Bunuel wrote:

The length of any side of a triangle must smaller than the sum of the other two sides. The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides.

Dear Bunuel, does this property apply to all quadrilaterals ? -- Thank you

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

Hi Bunnuel,

Is there a rule that pertains to the minimum length of polygons? I'm assuming that there is only one for triangles...

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?

gmatclubot

Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the
[#permalink]
25 May 2014, 13:08