A rectangular surface has length 4661 metres and breadth 3318 metres. On this area, square tiles are to be put. Find the maximum length of such tiles.

Given :

Length of the rectangular surface $= 4661 m$.

Breadth of the rectangular surface $= 3318 m$.

To find :

We have to find the Maximum length of the tiles.

Solution :

To find the maximum length of the tiles, we need to find the HCF of 4661 and 3318.

By Euclid's division lemma,

$$Dividend = Divisor \times Quotient + Remainder$$

Here, $4661 > 3318$.  

So, divide 4661 by 3318 

$4661 = 3318 \times 1 + 1343$

Remainder $= 1343$.

Repeat the above process until we will get 0 as the remainder.

Consider 3318 as the dividend and 1343 as the divisor,

$3318 = 1343 \times 2 + 632 $

Remainder $= 632$.

Consider 1343 as the dividend and 632 as the divisor,

$1343 = 632 \times 2 + 79$

Remainder $= 79$.

Consider 632 as the dividend and 79 as the divisor,

$632 = 79 \times 8 + 0$

Remainder $= 0$.

So, 79 is the Highest Common Divisor of 4661 and 3318.

Therefore,

The length of the square tile is 79 m. 

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Updated on: 10-Oct-2022

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