Find the least number of square tiles which can cover the floor of a rectangular shape having length and breadth 16 meters 58 centimetres and 8 meters 32 centimetres respectively.

Given :

Length of the rectangular floor $=$ 16 m 58 cm.

Breadth of the rectangular floor $=$ 8 m 32 cm.

To find :

We have to find the least number of square tiles which can cover the rectangular floor.

Solution :

Convert m to cm,

$1 m = 100 cm $

$16 m 58 cm = 16 \times 100 +58 cm = 1600 cm + 58 cm = 1658 cm$

 $8 m 32 cm = 8 \times 100 +32 cm = 800 cm + 32 cm = 832 cm$

 To find the least number of tiles, we need to find the side of largest square tile.

The highest common factor of 1658 and 832 is the side of largest square tile.

Prime factorization of $1658 = 2 \times 829 $

Prime factorization of $832 =  2 \times 2 \times 2 \times 109 = 2^3 \times 109$

HCF $=$ Smallest power of common prime factor.

HCF $=2$.

So, the side of the largest possible square tile $=2 cm$.

$Number of tiles required = \frac{Area of the rectangular floor}{Area of the tile}$   

Area of the rectangular floor $= 1658 \times 832 cm^2$

Area of the tile  $= 2 \times 2 = 4 cm^2$.

$Number of tiles required = \frac{1658 \times 832 cm^2}{4 cm^2}$

                                             $= 344864$.

Therefore, the least number of square tiles which can cover the rectangular floor is 344864.

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Updated on: 10-Oct-2022

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