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?
Answer & explanation
Correct answer: option 2
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.