One stands for one digit and two stands for two digits.

These two high figures are multiples of one digit.

1, which does not meet the requirement that the number of side length units in the problem is greater than 1.

3, the other four numbers are 1, 5, 7, 9, and the total is 22, so we can know that it is impossible to form a multiple of 3, so we will discard it.

5, the other four numbers are 1, 3, 7, 9, without 5 or 0, it is impossible to form a multiple of 5.

9, the other four numbers are 1, 3, 5, 7, and the total is 16. It is impossible to form a multiple of 3, let alone a multiple of 9.

There are only seven, and the other four numbers are 1, 3, 5, 7. It can be found that there can be two multiples of 7, 35 and 9 1.

So the side length of a square is 7, the length of a rectangle is 9 1 and the width is 35.