If the size of a tile is 9" by 9", how many tiles are required to cover a 12 ft. wide and 18 ft. long floor?

384

According to the question:

Area of the floor = 12 x 18 = 216 sq ft or 31,104 sq inch (as 1 ft = 12")

Area of the tile with size 9" x 9" = 81 sq inch

Number of tiles required to cover the area = \(\frac{31,104}{ 81}\) = 384 tiles

Thus, 384 tiles are required to cover a floor with sides 12 ft and 18 ft.