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

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13 Jan 2010, 14:52

Nothing strikes to my mind really. No formula, no fundamentals, absolutely nothing. All I can guess is 5 and 10 are reasonably closer to 2 + 4 than 15 so C. _________________

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

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13 Jan 2010, 17:55

Lets wait for Bunuel to look in to this question . I am not sure how to even solve this ? This is not a regular pentagon , angels are also not mentioned _________________

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

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13 Jan 2010, 18:22

Hi Here is my approach.. Split the pentagon into triangles. Properties of triangle with sides a,b and c. Sum of any two sides is greater than third side.( eg. a+b > c) Third side is greater than difference of the other two sides.( eg. a> b-c)

In triangle PQR PR can take values >1 and <5. So PR can be 2,3,4(considering only whole numbers based on the answer choices) In traingle RST, RT can be >1 and < 9. RT can be 2,3,4...8 Now for triangle PRT, consider exteme values of PR and RT to determine range of PT.

When PR=2, RT=2, PT is >0 and < 4. PT can be 1,2,3,4 When PR=4,RT=2, PT is >2 and <6. PT can be 3,4,5 When PR=2, RT =8, PT is >6 and <10. PT can be 7,8,9 When PR=4, RT=8, PT is >4 and < 12. PT can be 5,6,7...11

From all the values above and choices provided, PT can be 5 and 10. 15 is not possible. Hence IMO C.

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

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14 Jan 2010, 02:01

acegre wrote:

Hi Here is my approach.. Split the pentagon into triangles. Properties of triangle with sides a,b and c. Sum of any two sides is greater than third side.( eg. a+b > c) Third side is greater than difference of the other two sides.( eg. a> b-c)

In triangle PQR PR can take values >1 and <5. So PR can be 2,3,4(considering only whole numbers based on the answer choices) In traingle RST, RT can be >1 and < 9. RT can be 2,3,4...8 Now for triangle PRT, consider exteme values of PR and RT to determine range of PT.

When PR=2, RT=2, PT is >0 and < 4. PT can be 1,2,3,4 When PR=4,RT=2, PT is >2 and <6. PT can be 3,4,5 When PR=2, RT =8, PT is >6 and <10. PT can be 7,8,9 When PR=4, RT=8, PT is >4 and < 12. PT can be 5,6,7...11

From all the values above and choices provided, PT can be 5 and 10. 15 is not possible. Hence IMO C.

nice explanation.i have used the similar approach.but its still time consuming.

after going through this problem again.i thought of another approach.dont know whether its right.

we know for any triangle with side a,b,c

a-b < c < a+b

Consider triangle RST RS=4,ST=5, so 5-4<RT<5+4 ,1<RT<9

hi why i have ans C is... look the figure can be transformed to any shape as angles are not given... but the catch is PT cannot be greater than sum of all sides, sum is 14.. so 14 can be max ,... so 15 is out... however < 14 can be made with putting different angles between lines.. time reqd for coming to ans is less than 30 secs.. _________________

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

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14 Jan 2010, 03:29

2

This post received KUDOS

hrish88 wrote:

acegre wrote:

Hi Here is my approach.. Split the pentagon into triangles. Properties of triangle with sides a,b and c. Sum of any two sides is greater than third side.( eg. a+b > c) Third side is greater than difference of the other two sides.( eg. a> b-c)

In triangle PQR PR can take values >1 and <5. So PR can be 2,3,4(considering only whole numbers based on the answer choices) In traingle RST, RT can be >1 and < 9. RT can be 2,3,4...8 Now for triangle PRT, consider exteme values of PR and RT to determine range of PT.

When PR=2, RT=2, PT is >0 and < 4. PT can be 1,2,3,4 When PR=4,RT=2, PT is >2 and <6. PT can be 3,4,5 When PR=2, RT =8, PT is >6 and <10. PT can be 7,8,9 When PR=4, RT=8, PT is >4 and < 12. PT can be 5,6,7...11

From all the values above and choices provided, PT can be 5 and 10. 15 is not possible. Hence IMO C.

nice explanation.i have used the similar approach.but its still time consuming.

after going through this problem again.i thought of another approach.dont know whether its right.

we know for any triangle with side a,b,c

a-b < c < a+b

Consider triangle RST RS=4,ST=5, so 5-4<RT<5+4 ,1<RT<9

similarly for triangle QRT QR=2,1<RT<9

so 0<QT<10

finally for triangle PQT PQ=3,0<QT<10

2<PT<12.

Actually acegre's answer is really good. I tried to solve the problem and it only takes about a minute to solve. I solved it fast by just drawing everything and breaking down the pentagon into triangles. Please look at my picture and try to follow the logic. First, you must remember that in triangles one side should not be bigger or equal to the sum of two other sides. Then just draw small triangles and you will arrive to the correct answer, which is C. The PT should be less than 14. Thus the only possible answers are 5 and 10.

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

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14 Jan 2010, 05:20

vladik210 wrote:

hrish88 wrote:

acegre wrote:

Hi Here is my approach.. Split the pentagon into triangles. Properties of triangle with sides a,b and c. Sum of any two sides is greater than third side.( eg. a+b > c) Third side is greater than difference of the other two sides.( eg. a> b-c)

In triangle PQR PR can take values >1 and <5. So PR can be 2,3,4(considering only whole numbers based on the answer choices) In traingle RST, RT can be >1 and < 9. RT can be 2,3,4...8 Now for triangle PRT, consider exteme values of PR and RT to determine range of PT.

When PR=2, RT=2, PT is >0 and < 4. PT can be 1,2,3,4 When PR=4,RT=2, PT is >2 and <6. PT can be 3,4,5 When PR=2, RT =8, PT is >6 and <10. PT can be 7,8,9 When PR=4, RT=8, PT is >4 and < 12. PT can be 5,6,7...11

From all the values above and choices provided, PT can be 5 and 10. 15 is not possible. Hence IMO C.

nice explanation.i have used the similar approach.but its still time consuming.

after going through this problem again.i thought of another approach.dont know whether its right.

we know for any triangle with side a,b,c

a-b < c < a+b

Consider triangle RST RS=4,ST=5, so 5-4<RT<5+4 ,1<RT<9

similarly for triangle QRT QR=2,1<RT<9

so 0<QT<10

finally for triangle PQT PQ=3,0<QT<10

2<PT<12.

Actually acegre's answer is really good. I tried to solve the problem and it only takes about a minute to solve. I solved it fast by just drawing everything and breaking down the pentagon into triangles. Please look at my picture and try to follow the logic. First, you must remember that in triangles one side should not be bigger or equal to the sum of two other sides. Then just draw small triangles and you will arrive to the correct answer, which is C. The PT should be less than 14. Thus the only possible answers are 5 and 10.

u forgot to apply a-b<c here i.e one side should not be smaller or equal than the difference of two other sides

yeah acegre method is very good.i have used the same method.but its time consuming.

anyways my second method took less than 1 min.

chetan2u wrote:

chetan2u wrote:

is it C

hi why i have ans C is... look the figure can be transformed to any shape as angles are not given... but the catch is PT cannot be greater than sum of all sides, sum is 14.. so 14 can be max ,... so 15 is out... however < 14 can be made with putting different angles between lines.. time reqd for coming to ans is less than 30 secs..

very nice explanation Chetan.

can you elaborate on this however < 14 can be made with putting different angles between lines

hi hrish.. just take a stick 14m long which has one end as P and other as T..... other pts are Q, R ,S at 3m,5m and 9m resp from P...the stick is bent at these pts... first make it a straight line PT becomes 14m ,so 15m is out... second imagine putting T on top of P, it will become quad with PT as 0... take a pt 1m from P and im sure u can place T on top of it, here PT becomes 1m... lly u can place T at a dist of 5 and 10m also from P..... as u see the interior angles will adjust to give u the REQUIRED length of PT.. this is what i meant from however < 14 can be made with putting different angles between lines " _________________

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

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26 Mar 2010, 07:30

oh, cool.

I too tried to find an alternate explanation for this specific problem. I was unsure how to search for it. I have tried to solve this problem twice, and both times it took me more than 3 minutes.

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

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12 Feb 2016, 19:42

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