Perimeter question - simple way to teach
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Perimeter question - simple way to teach
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.
Re: Perimeter question - simple way to teach
What arrangement will give a small perimeter?
Long thin strip or a shape nearest a square?
If you can think about this then an answer will be easier to get.
Long thin strip or a shape nearest a square?
If you can think about this then an answer will be easier to get.
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Re: Perimeter question - simple way to teach
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
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 9:53 pm, edited 2 times in total.
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Re: Perimeter question - simple way to teach
Buy a book of squared paper...you will use it for other things too over the years.
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Re: Perimeter question - simple way to teach
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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Re: Perimeter question - simple way to teach
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.
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.
Re: Perimeter question - simple way to teach
Ah yes, but I can visualise things while I eat marshmallows that I cannot normally visualise. They definitely boost the power of the mind.
Re: Perimeter question - simple way to teach
Simple and easy to understand.Thanks a lot!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 :
No need to buy squared paper. Buy marshmallow instead
- 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