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Given that the length of each side of a quadrilateral is a distinct

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Given that the length of each side of a quadrilateral is a distinct [#permalink] New post   02 May 2018, 08:33
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Given that the length of each side of a quadrilateral is a distinct integer and that the longest side is not greater than 7,How many different possible combinations of side lengths are there?

[A] 21
[B] 24
[C] 32
[D] 34
[E] 35
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C
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Re: Given that the length of each side of a quadrilateral is a distinct [#permalink] New post   02 May 2018, 09:02
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That's really a nice question:

The concept used :

like the triangle inequality theorem, the sum of the lengths of the shortest three sides of a quadrilateral must be longer than the longest side.



Since the length of each side of a quadrilateral is a distinct integer and that the longest side is not greater than 7, all four sides must be taken from the set {1, 2, 3, 4, 5, 6, 7}. How many different sets of four could be selected? \(7C4\)= 7*6*5/(3!) = 7*5 = 35 Of those 35, the only ones that don’t work for sides of a quadrilateral are the ones in which the sum of the three smallest sides are equal to or less than the longest side. These would be

1 + 2 + 3 < 7

1 + 2 + 3 = 6

1 + 2 + 4 = 7

Those are the only invalid combinations. The other 32 work.

Answer = (C)
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Given that the length of each side of a quadrilateral is a distinct [#permalink] New post   02 May 2018, 09:13
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The simple proof of the concept used in this question :

"The sum of the lengths of the shortest three sides of a quadrilateral must be longer than the longest side."


Construct a random quadrilateral ABCD.
Attachment:
gmatbusters2.jpg
gmatbusters2.jpg [ 12.4 KiB | Viewed 15864 times ]

Join any two vertices(Say AC)

Now from triangle inequality j+k>n
Also n+m>l

So, from the above two equations we have j+k+m > l

Hence,

The sum of the lengths of the shortest three sides of a quadrilateral must be longer than the longest side.


It is analogous to triangle inequality theorem, which states that sum of two sides of a triangle is greater than third side.
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Re: Given that the length of each side of a quadrilateral is a distinct [#permalink] New post   11 Jun 2018, 23:03
Hi,

Sorry but why have you factored 1 + 2 + 3 two times? IMO the only combinations that dont satisfy the rule are 1,2,3 and 1,2,4. Thus 35 - 2 = 33 possible combinations.





gmatbusters wrote:
That's really a nice question:

The concept used :

like the triangle inequality theorem, the sum of the lengths of the shortest three sides of a quadrilateral must be longer than the longest side.



Since the length of each side of a quadrilateral is a distinct integer and that the longest side is not greater than 7, all four sides must be taken from the set {1, 2, 3, 4, 5, 6, 7}. How many different sets of four could be selected? \(7C4\)= 7*6*5/(3!) = 7*5 = 35 Of those 35, the only ones that don’t work for sides of a quadrilateral are the ones in which the sum of the three smallest sides are equal to or less than the longest side. These would be

1 + 2 + 3 < 7

1 + 2 + 3 = 6

1 + 2 + 4 = 7

Those are the only invalid combinations. The other 32 work.

Answer = (C)
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Re: Given that the length of each side of a quadrilateral is a distinct [#permalink] New post   12 Jun 2018, 00:29
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The three sets of side lengths that do not work are:

{1, 2, 3, 6}

{1, 2, 3, 7}

{1, 2, 4, 7}

For each of these sets, the sum of the 3 smallest sides is NOT greater than the length of the longest side. Therefore, these 3 side lengths don’t work.

ColonelRed wrote:
Hi,

Sorry but why have you factored 1 + 2 + 3 two times? IMO the only combinations that dont satisfy the rule are 1,2,3 and 1,2,4. Thus 35 - 2 = 33 possible combinations.





gmatbusters wrote:
That's really a nice question:

The concept used :

like the triangle inequality theorem, the sum of the lengths of the shortest three sides of a quadrilateral must be longer than the longest side.



Since the length of each side of a quadrilateral is a distinct integer and that the longest side is not greater than 7, all four sides must be taken from the set {1, 2, 3, 4, 5, 6, 7}. How many different sets of four could be selected? \(7C4\)= 7*6*5/(3!) = 7*5 = 35 Of those 35, the only ones that don’t work for sides of a quadrilateral are the ones in which the sum of the three smallest sides are equal to or less than the longest side. These would be

1 + 2 + 3 < 7

1 + 2 + 3 = 6

1 + 2 + 4 = 7

Those are the only invalid combinations. The other 32 work.

Answer = (C)


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Re: Given that the length of each side of a quadrilateral is a distinct [#permalink] New post   05 May 2020, 05:07
The sum of the lengths of the shortest three sides should always be greater than the longest side in a quadrilateral...
We are given that the lengths are integers 1,2,3...
If we take the sum of the minimum lengths 1,2&3, they add up to 6. So the fourth side cannot be 6 or 7.
So {1,2,3,6}& {1,2,3,7} not possible
Next, since the longest side is given to be 7 and if we add 1,2&4 they add up to 7, the combination {1,2,4&7} also won't work
Overall the number of combinations is 7C4=35
Out of this, as explained above 3 combinations won't work. So 35-3=32 is the answer
Hence C

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Re: Given that the length of each side of a quadrilateral is a distinct [#permalink] New post   29 Nov 2022, 22:22
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