A flooring tile has a shape of parallelogram whose base is $18 c m$ and the corresponding height is $6 c m$ . How many such tiles are required to cover a floor of area $540 m^{2} .$ (If required you can split the tiles in whatever way you want to fill up the corners)

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Solution

Given,
Base of the parallelogram - shaped flooring tile $= 18 c m$ and its height $= 6 c m$
So,
Area of one tile $= B a s e \times h e i g h t = 18 \times 6 = 108 {c m}^{2}$
We have the area of floor $= 540 m^{2}$
Hence, number of tiles $=$ Total area / Area of one tile
$= \frac{(540 \times 100 \times 100)}{108}$ $[∵ 1 m^{2} = (100 \times 100) {c m}^{2}]$
$= 50000$

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A flooring tile has a shape of parallelogram whose base is 18cm and the corresponding height is 6cm . How many such tiles are required to cover a floor of area 540m2. (If required you can split the tiles in whatever way you want to fill up the corners)

Given,Base of the parallelogram - shaped flooring tile =18cm and its height =6cmSo,Area of one tile =Base×height=18×6=108cm2We have the area of floor =540m2Hence, number of tiles = Total area / Area of one tile=(540×100×100)108 [∵1m2=(100×100)cm2]=50000

A flooring tile has a shape of parallelogram whose base is $18 c m$ and the corresponding height is $6 c m$ . How many such tiles are required to cover a floor of area $540 m^{2} .$ (If required you can split the tiles in whatever way you want to fill up the corners)