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# Perimeter Expressions

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.

Click here 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 rectangles.

Daniel also considered one of the later questions:

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 "b".

Benjamin from Wilson's School agreed:

I noticed that whatever way you arrange the two rectangles, the perimeter always equals:

(the longest length + the longest width) x 2

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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.

Click here 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 rectangles.

Daniel also considered one of the later questions:

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 "b".

Benjamin from Wilson's School agreed:

I noticed that whatever way you arrange the two rectangles, the perimeter always equals:

(the longest length + the longest width) x 2

How many solutions can you find to this sum? Each of the different letters stands for a different number.

Think of two whole numbers under 10, and follow the steps. I can work out both your numbers very quickly. How?

My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?