0.25 * √2 * A = 0.35355A. So it’s obvious that A = 3, width = 1.06066. ]]>

If the length of the square’s diagonal is \(d\), the rectangles have diagonals of length \(d\) and shorter sides of length \(\sqrt{2}(d/4)\). Hence their longer sides are of length \(\sqrt{14}(d/4)\). So we are looking for an integer length \(d\) such that the length of one of these sides is just larger than an inch.

1 2 3 4 5 6 7 8 9 |
d = 0 while True: d += 1 x, y = 2 ** (1/2) * d / 4, 14 ** (1/2) * d / 4 if 1.0 < x < 1.2 or 1.0 < y < 1.2: print(f"The square has a {d} inch diagonal(rectangles: {x:.3f}in x {y:.3f}in).") break |