A room is 8m×6m.how many tiles of 8cm×10cm are required to cover the floor?
The room is 800 cm by 600 cm, and the tile is 8 cm by 10 cm, so the tiles will fit exactly along either side. (ignoring any grouting gaps)
area of floor = 480,000 cm^2
area of one tile = 80 cm^2
number of tiles = 480,000/80 or 6000 tiles
or
tiles along the length = 800/8 = 100 tiles
tiles along the width = 600/10 = 60 tiles
number of tiles = 100(60) = 6000
or
area of floor = 8(6) or 48 m^2
area of one tile = .08(.10) or .006 m^2
number of tiles = 48/.006 or 6000 tiles
Area of the room=8m×6m
=48m²=48×10⁴cm
Area of tiles =8×10=80cm
No.of tiles=480000/80
=6000
To find the number of tiles required to cover the floor, we need to determine the total surface area of the floor and then divide it by the area of each tile.
First, let's convert the dimensions of the room into centimeters, as the dimensions of the tiles are given in centimeters.
The length of the room is 8m, which is equal to 800cm (since 1m = 100cm).
The width of the room is 6m, which is equal to 600cm.
Next, we need to calculate the area of the room in square centimeters:
Area of the room = length × width = 800cm × 600cm = 480,000 cm²
Now, let's calculate the area of each tile:
Area of the tile = length × width = 8cm × 10cm = 80 cm²
Finally, we can determine the number of tiles needed:
Number of tiles = Area of the room / Area of each tile = 480,000 cm² / 80 cm² = 6,000 tiles
Therefore, you would need a total of 6,000 tiles of size 8cm × 10cm to cover the entire floor.