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How many tiles of dimensions 12 cm and 5 cm will be needed to fit a region whose length and breadth are 144 cm and 100 cm.

Answer
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Hint: We will first of all write the formula for the area of the rectangle and then by putting in the given data, find the area of the tile and the region. After that, we will get the number of tiles by dividing the area of the region by the area of tiles.

Complete step-by-step answer:
We know that the area of rectangle is given by the following formula:-
\[ \Rightarrow Area = l \times b\], where l represents the length of the rectangle and b is the breadth of the rectangle.
Since, we are given that the tile is of dimensions 12 cm and 5 cm.
\[ \Rightarrow Area(tile) = (12 \times 5)c{m^2}\]
Simplifying the RHS, we will get:-
\[ \Rightarrow Area(tile) = 60c{m^2}\] …………..(1)
Now, we are also given the dimensions of the region which is 144 cm and 100 cm.
\[ \Rightarrow Area(region) = (144 \times 100)c{m^2}\]
Simplifying the RHS, we will get:-
\[ \Rightarrow Area(tile) = 14400c{m^2}\] …………..(2)
Now, we will find the number of tiles required to cover the region by dividing the area of the region by the area of tile.
$ \Rightarrow $ Number of tiles = $\dfrac{{14400}}{{60}}$
Cutting off the 0 from numerator and denominator, we will get:-
$ \Rightarrow $ Number of tiles = $\dfrac{{1440}}{6}$
Simplifying the calculations, we will get:-
$ \Rightarrow $ Number of tiles = 240

$\therefore $ We require 240 tiles to cover the region.

Note: The students must note that since we will fill the region using tiles and region filling requires the area of the region to be covered using the area of tiles, therefore, we used this concept of getting the number of tiles.
The students must note that we can relate the area of a rectangle to some real life examples. Like, suppose that we have 7 trees in each of the 10 rows. So, then in total we will have: 7 trees in each row and we have 10 rows, therefore, we need to add 7 10 times to get 70 trees in total, which is equivalent to finding the area of the rectangle with length as 7 units and breadth as 10 units.