Tricky Habs question - or maybe Im a bit tired...
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loveyouradvice
- Posts: 160
- Joined: Mon Nov 28, 2011 1:24 pm
Tricky Habs question - or maybe Im a bit tired...
here goes...
Shown various shapes and asked to fill in a grid of number of sides and number of diagonals
eg 3 sides, 0 diagonals
4 sides - 2 diagonals
5 - 5
6 - 9
7 - 12
8 - 20
n - ?
20 - ?
my daughter achieved the above - she could have slowly calculated 20 sided, but Im guessing they wanted you to work out n and calculate 20 sided from that
She spotted that n-1 and n sided shapes had a difference in number of diagonals as (n-2) but not the whole equation.
Love the formula nad how you reach it as a 10 year old!! huge thanks in advance - great admiration for hte brains on here!
Shown various shapes and asked to fill in a grid of number of sides and number of diagonals
eg 3 sides, 0 diagonals
4 sides - 2 diagonals
5 - 5
6 - 9
7 - 12
8 - 20
n - ?
20 - ?
my daughter achieved the above - she could have slowly calculated 20 sided, but Im guessing they wanted you to work out n and calculate 20 sided from that
She spotted that n-1 and n sided shapes had a difference in number of diagonals as (n-2) but not the whole equation.
Love the formula nad how you reach it as a 10 year old!! huge thanks in advance - great admiration for hte brains on here!
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hermanmunster
- Posts: 12901
- Joined: Fri Sep 15, 2006 9:51 am
- Location: The Seaside
Re: Tricky Habs question - or maybe Im a bit tired...
have googled "equation sides diagonals" and come up with quite a few pages showing the same equation:
n(n-3)/2
... however how you explain that to a 10 year old I don't know!
PS - not sure if 7-12 works??
7*4/2 = 14?
n(n-3)/2
... however how you explain that to a 10 year old I don't know!
PS - not sure if 7-12 works??
7*4/2 = 14?Re: Tricky Habs question - or maybe Im a bit tired...
Each vertex (corner) can have a diagonal to each of the other vertices EXCEPT itself, and it's two neighbours (because those would be sides, not diagonals).
Then multiply by the total number of vertices (which is the same as saying the number of sides).
But that would mean you were counting each diagonal twice - once from each end, so divide by 2
Therefore the number of diagonals = (number of sides) * (number of sides-3) / 2
or n(n-3)/2
(I knew I'd explained this one before - here - which is the same explanation in slightly different words.)
Then multiply by the total number of vertices (which is the same as saying the number of sides).
But that would mean you were counting each diagonal twice - once from each end, so divide by 2
Therefore the number of diagonals = (number of sides) * (number of sides-3) / 2
or n(n-3)/2
(I knew I'd explained this one before - here - which is the same explanation in slightly different words.)
Re: Tricky Habs question - or maybe Im a bit tired...
It's another example of these entrance tests being a bit unfair. The child who has done this before will quickly see it in an exam, the 10 year old child who hasn't but maybe could work it out for themselves won't have the time in an exam context unless they are extremely brilliant and very relaxed.
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loveyouradvice
- Posts: 160
- Joined: Mon Nov 28, 2011 1:24 pm
Re: Tricky Habs question - or maybe Im a bit tired...
Great answer - thank you!
And 100% agree... I think if I'd not been flustered, I could have worked it out - but probably taken me 7-10 mins which I cant imagine any 11 year old actually having spare in an exam! And as you said, they'd have had to be very relaxed....
So yes, only those who've done it before.... hmmmm!!!
And 100% agree... I think if I'd not been flustered, I could have worked it out - but probably taken me 7-10 mins which I cant imagine any 11 year old actually having spare in an exam! And as you said, they'd have had to be very relaxed....
So yes, only those who've done it before.... hmmmm!!!
