We had $60$ solutions submitted and a bit of response in the  blog   Here are a selection.

This solution was created by students at Primary and Secondary Master classes at the British School Al Khubairat in Abu Dhabi:

Solutions submitted by:
Olaf and James (Yr $6$), (Primary) $20$ mins including cutting time.
Thomas (Yr $7$), Joanna (Yr $9$) & Madeleine (Yr $9$), (Secondary) . $15$ mins including cutting time.

Dimensions of the Rectangle: 33 cm x 32 cm

Joe who studied this at home sent in the following:
Put nine, eight and fifteen along the top.  Place the $1$ under the bottom of the left corner on the eight.  Underneath the $9$ put the ten, next to the ten (and under the eight) put $7$.  This will line up the $7$ under the eight creating a place for $15$.  So fifteen is forming the top right hand corner of the rectangle.  Under the $7$ place $4$ so it is next to the bottom part of $10$.  Under the ten put fourteen , and next to the fourteen put eighteen. There you have it you have made a succseful rectangle with $33$cm long and $32$cm wide!!!!

From Kent College Prep School we had contributions from Megan, Polly and Amy, Sophia and Lottie, Olivia and Isabel  -  and Isabella and Victoria's solution was:  

Our first solution was we drew squares $1x1, 4x4, 7x7, 8x8, 9x9, 10x10, 14x14, 15x15$ and $18x18$.  Then we cut them out and started to find a solution.
It took us a couple of goes to find our solution; we found our solution because we started to put numbers together to make another number e.g. $8x8$ and $10x10$ made $18x18$.
When we had finished we counted the width and the length and it equalled $32x33$.

At the last moment we had a solution in from John at 37th of Heraklion School in
Greece:

We also had a solution from the British Manilla School in the Philippines sent in by Rishi.

We had a substantial contribution from year $6$ at East Ward School in Bury England. This can be seen here  .

A big WELL DONE to all those who sent in many other excellent correct solutions.