Question

A rectangular hall is 18m 72cm long and 13m 20cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.

Answer 1

Hint: Using the dimension of the rectangular hall, calculate its area. Assume a variable (say x) for length of side of square and a variable (say k) for the least number of required tiles. Prime factorise the area of the rectangular hall and obtain the largest “perfect square” which can be deduced from the area and then divide the area of the rectangular hall by the value of ${{x}^{2}}$ i.e. the largest perfect square obtained.

Complete step by step answer:
The rectangular hall is 18m 72cm long and 13m 20cm broad.
We know area of rectangle $=\left( length \right)\times \left( breadth \right)$
So, area of this rectangular hall $=\left( 18m72cm \right)\times \left( 13m20cm \right)$
We know, 1m = 100cm
So, area of this rectangular hall $=\left( 1872cm \right)\times \left( 1320cm \right)$
We have to cover this area with square tiles of the same area.
Let the length of the square tile used be $xcm$, so the area of the square tile $={{\left( side \right)}^{2}}$. [As area of a square $={{\left( side \right)}^{2}}$ ]
$\therefore $ Area of one square tile $={{x}^{2}}c{{m}^{2}}$
Let the number of tiles used be k
Thus, the ‘k’ tiles of area $\left( {{x}^{2}} \right)c{{m}^{2}}$ will be used to pave this hall of area $\left( 1872\times 1320c{{m}^{2}} \right)$.
So, the total area of k tiles = Area of the hall
i.e. $k{{x}^{2}}=\left( 1872\times 1320 \right)$
According to the question, we want the number of tiles to be least possible,
$k\times {{x}^{2}}=\left( 1872\times 1320 \right)$
For ‘k’ to be least possible $\left( {{x}^{2}} \right)$ should be maximum possible.
Let us prime factorise $\left( 1872\times 1320 \right)c{{m}^{2}}$:
\[\begin{align}
  & 2\left| \!{\underline {\,
  1872 \,}} \right. \\
 & 2\left| \!{\underline {\,
  936 \,}} \right. \\
 & 2\left| \!{\underline {\,
  468 \,}} \right. \\
 & 2\left| \!{\underline {\,
  234 \,}} \right. \\
 & 3\left| \!{\underline {\,
  117 \,}} \right. \\
 & 3\left| \!{\underline {\,
  39 \,}} \right. \\
 & \ \ \left| \!{\underline {\,
  13 \,}} \right. \\
\end{align}\]

$\begin{align}
  & 2\left| \!{\underline {\,
  1320 \,}} \right. \\
 & 2\left| \!{\underline {\,
  660 \,}} \right. \\
 & 2\left| \!{\underline {\,
  330 \,}} \right. \\
 & 3\left| \!{\underline {\,
  165 \,}} \right. \\
 & 5\left| \!{\underline {\,
  55 \,}} \right. \\
 & \ \,\left| \!{\underline {\,
  11 \,}} \right. \\
\end{align}$

Now, putting $\left( 1872\times 1320 \right)=\left( 2\times 2\times 2\times 2\times 3\times 3\times 13 \right)\times \left( 2\times 2\times 2\times 3\times 5\times 11 \right)$ in the equation above, we will get,
$k{{x}^{2}}=\left( 2\times 2\times 2\times 2\times 3\times 3\times 13 \right)\times \left( 2\times 2\times 2\times 3\times 5\times 11 \right)$
We have to find the maximum value of ${{x}^{2}}$, so from the RHS of the equation, we will find the greatest “perfect square number” possible.
\[k{{x}^{2}}={{2}^{7}}\times {{3}^{3}}\times {{5}^{1}}\times 11\times 13\]
\[k{{x}^{2}}={{2}^{6}}\times 2\times 3\times 5\times 11\times 13\]
\[\begin{align}
  & we\ know\ {{2}^{a+b}}={{2}^{a}}\times {{2}^{b}} \\
 & so,{{2}^{7}}={{2}^{6+1}}={{2}^{6}}\times 2 \\
 & {{3}^{3}}={{3}^{\left( 2+1 \right)}} \\
\end{align}\]
\[k{{x}^{2}}={{\left( {{2}^{3}} \right)}^{2}}\times {{3}^{2}}\times 2\times 3\times 5\times 11\times 13\]
Maximum perfect square which we can get is ${{\left( {{2}^{3}} \right)}^{2}}\times {{3}^{2}}$.
$\Rightarrow {{x}^{2}}={{\left( {{2}^{3}} \right)}^{2}}\times {{3}^{2}}$
Now putting ${{\left( {{2}^{3}} \right)}^{2}}\times {{3}^{2}}$ in place of ${{x}^{2}}$ in the above equation, we will get,
\[k{{\left( {{2}^{3}} \right)}^{2}}\times {{3}^{2}}={{\left( {{2}^{3}} \right)}^{2}}\times 2\times 3\times 5\times 11\times 13\]
On dividing both sides of equation by${{\left( {{2}^{3}} \right)}^{2}}\times {{3}^{2}}$, we will get,
\[\begin{align}
  & k=2\times 3\times 5\times 11\times 13 \\
 & \Rightarrow k=4290 \\
\end{align}\]

So, the least number of square tiles required to pave the given hall is 4290.

Note: Dimensions of hall are given in “x m y cm” form. First convert them into centimetres for further calculation. If you convert in meters, you won’t be able to prime factorise and won’t be able to get maximum value of area of square tiles.