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 Post subject: Perimeter question - simple way to teachPosted: Sat Nov 23, 2013 5:14 pm

Joined: Fri Oct 12, 2012 11:22 pm
Posts: 28
I have 42 square photographs each measuring 5cm by 5cm. How can I arrange them all into a rectangular shaped picture with the shortest possible length of picture frame. I need a straightforward way to show DC as finding the factors appears to be too much of a trial and error.

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 Post subject: Re: Perimeter question - simple way to teachPosted: Sat Nov 23, 2013 5:26 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 14010
What arrangement will give a small perimeter?

Long thin strip or a shape nearest a square?

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 Post subject: Re: Perimeter question - simple way to teachPosted: Sat Nov 23, 2013 7:02 pm

Joined: Thu Oct 18, 2012 9:41 am
Posts: 444
using Guest55's method:

if you work it out as a long and thin rectangle then 5 * 42 will give one length = 210, double it for the opposite side = 420 then the two remaining sides will be 10 = 430

If you work it out as a 6 * 7 rectangle then top and bottom sides will be 60 and left and right sides will be 70, gives total of 130

therefore shortest perimeter will be the 6*7 option

Last edited by SleepyHead on Sat Nov 23, 2013 8:53 pm, edited 2 times in total.

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 Post subject: Re: Perimeter question - simple way to teachPosted: Sat Nov 23, 2013 7:57 pm

Joined: Tue Dec 18, 2012 10:59 am
Posts: 3579
Buy a book of squared paper...you will use it for other things too over the years.

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 Post subject: Re: Perimeter question - simple way to teachPosted: Wed Nov 27, 2013 12:42 am

Joined: Mon Aug 12, 2013 9:13 am
Posts: 452
The length of square does not matter. Try doing it for a square 1 cm x 1 cm (and the answer will be same). It is the shortest of the combinations which are based on factors concept :

42 x 1 i.e. all 42 in one row : (42 x 2) + (1 x 2) = 86
21 x 2 i.e. two stacks of 21 each = (21 x 2) + (2 x 2) = 46
14 x 3 i.e. three stacks of 14 each = (14 x 2) + (3 x 2) = 34
7 x 6 i.e. seven stacks of 6 each = (7 x 2) + (6 x 2) = 26

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 Post subject: Re: Perimeter question - simple way to teachPosted: Wed Nov 27, 2013 5:55 am

Joined: Tue Dec 18, 2012 10:59 am
Posts: 3579
Unfortunately although marshmallows are great for multiplication and factors across the kitchen table, they will do little to help a 9 year old visualise the problem they have a mental block on calculating. Squared paper and a set of felt pens will hopefully help their mind equate the sum required to the problem set before them. Having a child who struggles with problem solving, I have learnt that making things tangible seems to be the ticket to less frowning.

Squared paper also helps with area problems, lines of symmetry, and working out tangram puzzles, if you do not wish to buy a whole pad there are plenty of web sites that allow you to print an accurate sheet.

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 Post subject: Re: Perimeter question - simple way to teachPosted: Wed Nov 27, 2013 9:32 am

Joined: Tue Jul 21, 2009 9:56 pm
Posts: 8551
Ah yes, but I can visualise things while I eat marshmallows that I cannot normally visualise. They definitely boost the power of the mind.

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 Post subject: Re: Perimeter question - simple way to teachPosted: Wed Nov 27, 2013 11:11 am

Joined: Wed Sep 11, 2013 12:28 pm
Posts: 20
parent2013 wrote:
The length of square does not matter. Try doing it for a square 1 cm x 1 cm (and the answer will be same). It is the shortest of the combinations which are based on factors concept :

42 x 1 i.e. all 42 in one row : (42 x 2) + (1 x 2) = 86
21 x 2 i.e. two stacks of 21 each = (21 x 2) + (2 x 2) = 46
14 x 3 i.e. three stacks of 14 each = (14 x 2) + (3 x 2) = 34
7 x 6 i.e. seven stacks of 6 each = (7 x 2) + (6 x 2) = 26

Simple and easy to understand.Thanks a lot!

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