Daniel from Staplehurst Primary School sent
us a very clear solution to this problem:
Alison's shape with a perimeter of 8a + 6b combined the largest and
smallest rectangles, with the smaller rectangle's side of length a
against one of the larger rectangle's sides.
I have made the largest possible perimeter, 12a + 12b, using
all the pieces because I have put the shorter side of each shape
against the one before it.
to see the examples and
all the working out.
Rajeev from Haberdashers' Aske's Boys'
School managed to improve on Daniel's solution for the largest
perimeter. Take a look here at how he combined the
Daniel also considered one of the later
I would like to see Charlie's second shape with a perimeter of 7a +
4b because I think he has made a mistake: it is impossible to find
a shape with an odd number of "a"s.
It is impossible to get an odd number of "a" or "b" on the
perimeter because every shape has an even number of "a"s or "b"s.
When you place the a against an existing shape you are taking away
an a from the perimeter and then adding one back again as well as
adding 2b and vice versa, this is true with all the rectangles that
you add on. Therefore you always get an even number of "a" or
Benjamin from Wilson's
I noticed that whatever way you arrange the two rectangles,
the perimeter always equals:
(the longest length + the longest width) x 2