Answer

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Hint: For the tiles to fit in the rectangular region, their combined area must be equal to the area of the rectangular region.

According to the question, the length and breadth of the tiles are $l = 12cm$ and $b = 5cm$ respectively.

And we know that the area of a rectangle is $A = l \times b$. So, using this formula we will find out the area of each tile. We’ll get:

$

\Rightarrow A = 12 \times 5, \\

\Rightarrow A = 60c{m^2} \\

$

The length and breadth of the rectangular region are $100cm$ and $144cm $respectively. So, its area will be:

$

\Rightarrow {A_R} = 100 \times 144, \\

\Rightarrow {A_R} = 14400 \\

$

Suppose we have $n$ number of tiles covering the entire region of the rectangle. Then the combined area of $n$ tiles must be the same as the area of the rectangle. So, we have:

$

\Rightarrow n \times 60 = 14400, \\

\Rightarrow n = \dfrac{{14400}}{{60}}, \\

\Rightarrow n = 240 \\

$

Therefore, we need $240$ tiles to fit in the entire rectangular region.

Note: Since the rectangle is a two dimensional figure, we have compared the area. For three dimensional figures such as cylinder, sphere or cone we always compare volume for such types of problems.

According to the question, the length and breadth of the tiles are $l = 12cm$ and $b = 5cm$ respectively.

And we know that the area of a rectangle is $A = l \times b$. So, using this formula we will find out the area of each tile. We’ll get:

$

\Rightarrow A = 12 \times 5, \\

\Rightarrow A = 60c{m^2} \\

$

The length and breadth of the rectangular region are $100cm$ and $144cm $respectively. So, its area will be:

$

\Rightarrow {A_R} = 100 \times 144, \\

\Rightarrow {A_R} = 14400 \\

$

Suppose we have $n$ number of tiles covering the entire region of the rectangle. Then the combined area of $n$ tiles must be the same as the area of the rectangle. So, we have:

$

\Rightarrow n \times 60 = 14400, \\

\Rightarrow n = \dfrac{{14400}}{{60}}, \\

\Rightarrow n = 240 \\

$

Therefore, we need $240$ tiles to fit in the entire rectangular region.

Note: Since the rectangle is a two dimensional figure, we have compared the area. For three dimensional figures such as cylinder, sphere or cone we always compare volume for such types of problems.

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