A floor is 4.5 m long and 3.6 m wide A 6 cm square tile costs ₹ 23.25.
What will be the cost to cover the floor with these tiles?

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Solution

Length of floor = 4.5 m and width of floor = 3.6 m
Area of floor = length $\times$ Breadth $= 4.5 \times 3.6 m^{2} [∵ w i d t h = b r e a d t h = 3.6 m]$
Side of square tile $= 6 c m = \frac{6}{100} = 0.06 m$
Area f one tile $= (s i d e)^{2} = 0.06 \times 0.06 m^{2}$
NUmber of tiles $\frac{A r e a o f f l o o r}{A r e a o f o n e t i l e} = \frac{4.5 \times 3.6}{0.06 \times 0.06} = 4500$
$∵$ cost fo 1 tile = ₹ 23.25
$∴$ cost of $4500 t i l e s ₹ 23.25 \times 4500 = ₹ 104625$
Hence,₹ 104625 are required to cover the floor with tiles .

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A floor is 4.5 m long and 3.6 m wide A 6 cm square tile costs ₹ 23.25. What will be the cost to cover the floor with these tiles?

A floor is 4.5 m long and 3.6 m wide A 6 cm square tile costs ₹ 23.25.
What will be the cost to cover the floor with these tiles?