How is 4 the maximum number of nonoverlapping squares with sides of length 3 that will fit inside of a square with sides of length 6?

Answer 1

Each of the four squares, combined, must have a perimeterwhich adds up to that of the larger square. Since the larger square has a perimeter of 24, so must these squares. However, since we are placing them next to each other, the 2 sides of each square facing other sides do not count. Each individual square has a perimeter of 12, so that makes 48. Subtracting the lengths of the sides placed adjacent to each other (8 sides x a length of 3 = 24), we get a perimeter of 24, the perimeter of the larger square. Because the perimeters are equal, there is no room left for another square without overlapping.