Trickier hand tiling

This should be a little trickier. Using $20$ rectangles of aspect ratio $1:2$, make two squares. Use sizes 2,3,7,12,16 in both squares. And split this list among the two squares: 6,8,10,11,13,15,17,18,20,22
In other words, the first square uses all five from the first list and five from the second. The second square uses the first list plus the other five from the second list. Sizes 2,3,7 means $2\times4$, $3\times6$, $7\times14$ etc

As usual, no gaps or overlaps.

A computer would spoil it. But it's probably fairly tough by hand.

@tom was correct about the size of the squares and the combinations.

I calculated the total area, plus the naive minimum and maximum dimensions of the squares by using the smallest 5 variable rectangles and largest 5 rectangles to get a range of 44 to 66, then checked x=44 to 56 to see if the total area minus x^2 was itself a square (57+ is symmetric to 44-56 so need not be checked). 56 was the only one that worked out, with the remainder being itself the square of 56. I didn't try out all the combinations of the 10 variable rectangles, but verified that the total area of one of the sets of 5 mentioned by tom was equal to half the total area of the full set.

I tiled them on graph paper shown below. Placing them roughly largest to smallest, trying to avoid leaving small gaps where possible.

• great job :-) !! – tom Feb 19 '18 at 16:54