Question

40
1) How many tiles of length 8cm
and width 5cm are needed to tile
up the floor of length 200cm and
width 200cm​

Answer 1

Answer:

$\red\bigstar$ The number of tiles that  are needed to tile up the floor = 1000.

$\dag$SOLUTION$\dag$

$\boxed{\sf Number \: of \; tiles\; needed \;for\; tiling\; of\; the\; floor = \dfrac{Area\; of\; the\; floor}{Area \;of \;the \;tile} }$

Area of the floor

The floor is in the shape of a square

Area of a square =  , Where a is the side of the square.

Here,

a = 200 cm.

DIAGRAM

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf }\put(-0.5,4.2){\bf }\put(4.2,-0.5){\bf }\put(4.2,4.2){\bf }\put(1.5,-0.6){\bf\large 200\ cm}\put(4.4,2){\bf\large 200\ cm}\end{picture}$

⇒ Area of the floor = (200)²

⇒ Area of the floor = 40000 cm².

Area of the tile

The tiles are  in the shape of a rectangle.

Area of a rectangle = l × b , Where l is the length and b is the breadth of the rectangle.

Here,

l = 8 cm.

b = 5 cm.

DIAGRAM

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 8 cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 5 cm}\put(-0.5,-0.4){\bf }\put(-0.5,3.2){\bf }\put(5.3,-0.4){\bf }\put(5.3,3.2){\bf }\end{picture}$

⇒ Area of the tile = 8 × 5

⇒ Area of the tile = 40 cm².

We know that,

$\sf Number \: of \; tiles\; needed \;for\; tiling\; of\; the\; floor = \dfrac{Area\; of\; the\; floor}{Area \;of \;the \;tile} }$

Here,

Area of the floor = 40000 cm².

Area of the  tile = 40 cm².

$\implies \sf Number \: of \; tiles\; needed \;for\; tiling\; of\; the\; floor = \dfrac{40000}{40} }$

$\implies \sf Number \: of \; tiles\; needed \;for\; tiling\; of\; the\; floor = 1000 }$