The 12 -pieces, covering 60 squares, can be arranged to form rectangles of any of the shapes 3×20, 4×15, 5×12 or 6×10. The earliest solutions to these are shown.
3×20: F. Hansson (PFCS Jun/Aug 1935, problem ¶1844), gave the only two cases possible; a result confirmed by D. S. Scott with computer in 1958. For the second solution rotate the marked section.
4×15: W. E. Lester (PFCS Feb/Apr 1935 ¶1679), where every piece abuts an edge. W. Stead (FCR Dec 1954 ¶21) gave the second example.
5×12: W. E. Lester (PFCS Feb/Apr 1935 ¶1680, April/June 1935 ¶1780), in the second every piece abuts an edge. W. Stead (FCR Dec 1954 ¶20) gives an example divided 5 and 7.
6×10: W. E. Lester (PFCS Feb/Apr 1935 ¶1681, Apr/Jun 1935 ¶1781-2) gave the first three: in the second every piece abuts an edge; in the third the 1×5 piece is internal.
The fourth, by T. R. Dawson (FCR Aug/Oct 1948 ¶7792), and the fifth by. H. D. Benjamin (FCR Aug/Feb 1950/1 ¶8794) were composed as examples of 'degenerate half-hollow rectangles' of width 5 (having a unit cut in the middle of the 10-unit edge).
F. Hansson (FCR Oct/Feb 1954/5 ¶10,057) made the four examples with 'crossroads' in symmetric positions.
There are, according to Prof. C. B. Haselgrove, 2339 solutions to the 6×10 problem.
An H-shape for happy birthday by R. J. French, W. E. Lester, H. Perkins and W. J. Tibbs (FCR Jun/Aug 1944 ¶6016). The two right-hand files can be turned upside down. "This board of composers was self-constituted while solving No. 4142".
T. R. Dawson (FCR Aug/Oct 1946 ¶6884). The s forming a 7×9 hollow rectangle with 1×3 hole. The rectangle 6×10, having a slit in the middle, can be regarded as a limiting case of a 'hollow rectangle'. See also under Multiple Shape Constructions below.
W. Stead (FCR Dec 1954, ¶18, ¶24) 7×9 and 3×21 rectangles with 3 unit holes equispaced.
Square 9×9 minus triangle by H. D. Benjamin (FCR Dec/Feb 1945/6 ¶6628).
Half-hollow rectangles by H. D. Benjamin (FCR Aug/Feb 1950/1 ¶8793) width 4, 7×9, 6×11, 5×13 and by T. R. Dawson (FCR Aug/Oct 1948 ¶7792) another 6×11 width 4 and one 4×18, width 3.
For 5-width degenerate examples see the 6×10 rectangles above. See also under Multiple Shape Constructions below for further examples (bisected half-hollow rectangles).
H. D. Benjamin (FCR Dec/Apr 1939/40 ¶4142). An H-shape for Happy Birthday (or Hurdle?). For a more attractive result see the biaxially symmetric version above.
Chess rook by H. D. Benjamin (FCR Feb/Apr 1939 ¶3569).
A (rather sinister?) monument by O. Weisert (FCR Dec/Jun 1939/40 ¶4151).
H. D. Benjamin (FCR Aug/Oct 1938 ¶3305). An L-shape, 8×11, with every piece on edge. The remaining 4×7 area could be filled with the  to  pieces.
W. Stead (FCR Apr/Aug 1953 ¶9543). Teapot.
W. Stead (FCR Aug/Dec 1954 ¶10,003). Key.
A 'Christmas Castle' by Dorothy Dawson (FCR Dec/Feb 1936/7 ¶2558). Three solvers sent answers with door and loopholes placed symmetrically.
Two L-shaped pieces. T. R. Dawson (FCR Aug/Oct 1946 ¶6884). Ls, width 3, 6×7. H. D. Benjamin (FCR Aug/Oct 1949 ¶8223) Ls, width 3, all possible cases, 4×9, 5×8, 6×7. P. B. van Dalfsen (FCR Jun/Oct 1953 ¶9602) L, width 3, 5×8.
Depending on their proportions, these L-shapes can be combined together to form rectangles, or hollow rectangles or half-hollow rectangles in various ways. For example, the 6×7 pairs can form either a 6×10 (with 0×4 central cut) or a 7×9 with 1×3 hole. The 5×8 pairs combine to form an 8×8 square with 2×2 hole. The 4×9 pair combines to form an 8×9 with 1×3 hole. The Benjamin results were originally presented as 'bisected half-hollow-rectangles' forming two different types of these. The Dalfsen result was presented as a simple chessboard dissection.
W. E. Lester and B. Zastrow, independently in solving (PFCS Aug/Oct 1935 ¶1923) covered an 8×8 board, minus corners, by using these two components. They can also be combined to cover the 8×8 board minus two dominos, cut from the middle of opposite edges.
R. J. French (FCR Oct/Dec 1945 ¶6575) asked: What cross-section has a box which exactly contains the 5-square pieces packed in 3 layers? He gave the first solution. The second is by W. H. Reilly. W. Stead (FCR Oct/Feb 1954/5 ¶10,067) later added the two other cases on the right, one with a hole.