Question

A hall is rectangular in shape of length 25 m and breadth 18 m. its floor meer to be
paved with square bles of 25 cm. How many such tiles are required to cover the cover of the hall​

Answer 1

Answer:

18×25=450

450÷25=18tiles

Answer 2

Analysis:-

‎ ‎ ‎ ‎ ‎ ‎ ‎We're given with the dimensions of the floor of a rectangular hall. And we're asked to find the number of tiles of a particular shape and dimension, required to pave its floor.

Dimensions of the floor: Length = 25 m and Breadth = 18 m.

Shape and dimension(s) of the tile: It is in the shape of square with side = 25 cm.

Understanding the concept:-

‎ ‎ ‎ ‎ ‎ ‎ ‎ This question is from the chapter "Mensuration" which is branch of Mathematics that deals with the computation of geometric magnitudes, such as the the perimeter of a surface, the area of a surface and the volume of a solid.

‎ ‎ ‎ ‎ ‎ ‎ The geometric figures focused here are ‎rectangle and square. Rectangle is a four sided polygonal figure, i.e, a quadrilateral, in which the opposite sides are of equal length and are parallel. Also its diagonals bisects each other.

$.\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge \boxed{ \sf{ \: \: \: \: \: \: }} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tiny\sf{A \: rectangle}$

Perimeter = 2 (l + b) units

Area = lb sq.units

A square is also quadrilateral, in which all the four sides are equal, as a result, it's diagonals bisects at 90°.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge \boxed{ \sf{ \:}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tiny\sf{A \: square}$

Perimeter = 4s units

Area = s × s (or) s² sq.units

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Solution:-

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎As per the analysis, we need to find the number of tiles required to pave the floor of the hall. How can we find it? We can find it using the topic - "Area of a rectangle and area of a square".

To find the required answer, we've to first find the area of the floor and the area of a tile. And then, we have to divide the area of floor by area of a tile, which is the required answer.

Before finding the areas, check whether the units of the measures of both the figures are same! Here, it isn't same. So, we have to convert the units from "metres" to "centimetres".

As we know, $\boxed{ \sf{1 \: m = 100 \: cm}}$,

$\\ \longmapsto{ \sf{25 \: m \: (length) = 25 \times 100 = 2500 \: cm}} \\ \\ \longmapsto{ \sf{18 \: m \: (breadth) = 18 \times 100 = 1800 \: cm}}$

Now, let's find the area of the floor –

$\longmapsto{ \sf{ \pmb{Area \: _{(rectangle)} = lb \: sq.units}}} \\ \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2500 \times 1800} \\ \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \underline{4500000 \: {cm}^{2}} }$

Finding the area of a tile –

$\longmapsto{ \sf{ \pmb{Area \: _{(square)} = {(s)}^{2} \: sq.units}}} \\ \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {(25)}^{2} } \\ \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \underline{625 \: {cm}^{2}} }$

Finally, let's find the number of tiles required by substituting the obtained measures in:

$\longmapsto{ \sf{ \dfrac{Area \: of \: floor \: of \: the \: hall}{Area \: of \: a \: tile} }} \\ \\ \longmapsto{ \sf{ \dfrac{ \cancel{4500000}}{ \cancel{625}}}} \\ \\ \longmapsto \underline{\boxed{ \tt{ \pmb{ \red{7200}}}}}$

$\\ \therefore \underline{ \sf{ \pmb{ \red{7200 \: tiles} \: are \: required \: to \: pave \: the \: floor .}}}$

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