401) How many tiles of length 8cmand width 5cm are needed to tileup the floor of length 200cm andwidth 200cm About the author Claire
Answer: [tex]\red\bigstar[/tex] The number of tiles that are needed to tile up the floor = 1000. [tex]\dag[/tex]SOLUTION[tex]\dag[/tex] [tex]\boxed{\sf Number \: of \; tiles\; needed \;for\; tiling\; of\; the\; floor = \dfrac{Area\; of\; the\; floor}{Area \;of \;the \;tile} }[/tex] 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 [tex]\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}[/tex] ⇒ 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 [tex]\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}[/tex] ⇒ Area of the tile = 8 × 5 ⇒ Area of the tile = 40 cm². We know that, [tex]\sf Number \: of \; tiles\; needed \;for\; tiling\; of\; the\; floor = \dfrac{Area\; of\; the\; floor}{Area \;of \;the \;tile} }[/tex] Here, Area of the floor = 40000 cm². Area of the tile = 40 cm². [tex]\implies \sf Number \: of \; tiles\; needed \;for\; tiling\; of\; the\; floor = \dfrac{40000}{40} }[/tex] [tex]\implies \sf Number \: of \; tiles\; needed \;for\; tiling\; of\; the\; floor = 1000 }[/tex] Reply
Answer:
[tex]\red\bigstar[/tex] The number of tiles that are needed to tile up the floor = 1000.
[tex]\dag[/tex]SOLUTION[tex]\dag[/tex]
[tex]\boxed{\sf Number \: of \; tiles\; needed \;for\; tiling\; of\; the\; floor = \dfrac{Area\; of\; the\; floor}{Area \;of \;the \;tile} }[/tex]
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
[tex]\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}[/tex]
⇒ 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
[tex]\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}[/tex]
⇒ Area of the tile = 8 × 5
⇒ Area of the tile = 40 cm².
We know that,
[tex]\sf Number \: of \; tiles\; needed \;for\; tiling\; of\; the\; floor = \dfrac{Area\; of\; the\; floor}{Area \;of \;the \;tile} }[/tex]
Here,
Area of the floor = 40000 cm².
Area of the tile = 40 cm².
[tex]\implies \sf Number \: of \; tiles\; needed \;for\; tiling\; of\; the\; floor = \dfrac{40000}{40} }[/tex]
[tex]\implies \sf Number \: of \; tiles\; needed \;for\; tiling\; of\; the\; floor = 1000 }[/tex]