Question

A warehouse's rectangular ceiling measures 12 meters by 18 meters, and is covered by acoustic tile squares which are one meter on a side. Each piece of tile weighs a quarter of a kilogram. What will be the total weight of the tile required to cover the ceiling?

Answer 1

To cover a 12m x 18m ceiling with 1m² acoustic tiles weighing 0.25kg each, we need 216 tiles, resulting in a total weight of 54 kilograms for the tile coverage.

Step-by-step explanation:

The student is tasked with finding the total weight of the acoustic tile squares needed to cover a warehouse ceiling that measures 12 meters by 18 meters. Each tile square is one meter on a side and weighs a quarter of a kilogram.

To solve this, we first calculate the total area of the ceiling by multiplying its length by its width:

12 meters * 18 meters = 216 square meters.

Since each acoustic tile covers one square meter, we will need 216 tiles to cover the entire ceiling.

Next, we calculate the total weight:

216 tiles * 0.25 kilograms per tile = 54 kilograms.

Therefore, the total weight of the tiles required to cover the ceiling is 54 kilograms.

Answer 2

The first thing to do is find the rectangular warehouse area.
We have then:
A = (12) * (18) = 216
A = 216 m ^ 2
We now look for the area of each acoustic tile square:
A '= (1) ^ 2 = 1
A '= 1 m ^ 2
We look for acoustic tile squares number:
N = (A) / (A ')
N = (216) / (1)
N = 216
Finally we look for the total weight:
W = (1/4) * (216)
W = 54 Kg
Answer:
The total weight of the tile required to cover the ceiling is:
W = 54 Kg