1) calculate the total area of the quilt adding the single areas of all 9 squares ( 1+16+49+64+81+100+196+225+324= 1056 cm2)
2) calculate the Square Ruth of 1056 in order to obtain a close indication of a possible side of the rectangle as squares can be considered rectangles with all equal sides. I obtained 32.4961. I rounded up to 32.5 cm.
The total perimeter is P=32.5x4= 130 cm
3) Out of my 9 initial squares, none has a side with decimal, so I had to redistribute my 0.5 cm in order to obtain a rectangle. So I supposed to have 1 side of 32cm and the other of 33 cm. The perimeter and the area are the same
P= 2(32+33)
P= 2x65 = 130 cm
A= 32x33 = 1056 cm2
4) Out of my lengths squares sides (1,4,7,8,9,10,14,15 & 18) which is the easiest sum combination that can give me 32 and 33?
I considered that the small square of 1 cm2 for sure will be somewhere in the middle, otherwise I can't build up a rectangle and most probably also the 4 cm2 one will stay in the middle, so for now I don't consider these 2 squares.
The easiest sum to get to 33 is 18+15.
The other side of 32 cm needs to have one side of a smaller square in common, so on one side I have to reach 32, starting from 18 cm (32-18=14), on the opposite side I have to reach 32, starting from 15 cm (32-15=17).
I have available a square of 14 cm, but not one with a side of 17 cm.
With the error and trial method, I considered the possible combinations of 17 (10+7 or 8+9) and tried to build up the rectangle. In the side of 32, I can't have the 10 cm square next to the 15cm square, otherwise then I would need a square of 5 cm to close the gap above (15-10=5) and I don't have it. So for sure I need the 8 cm and 9 cm squares.
On the 33 cm side, I already have the 14 cm square and I just concluded that I have to put there the 10 cm square too. 14+10=24 and 33-24=9. I found out the correct position of the 9 cm square.
Now the only 3 squares left over are: 1,4,7
Under the 14 cm square I put the 10 cm square creating a perfect gap of 4 cm to connect with the 18 cm square.
Under the 4 cm square, I can put the 7 cm one (4+18=22 and 22-15=7).
And finally the last square of 1 cm to close the last gap. (see picture attached done in xls to show the result).