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  1. Chapter 10 Class 6 Mensuration
  2. Serial order wise

Transcript

Ex 10.3, 12 How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively: (a) 100 cm and 144 cm Here, Number of tiles = (๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘Ÿ๐‘’๐‘๐‘ก๐‘Ž๐‘›๐‘”๐‘ข๐‘™๐‘Ž๐‘Ÿ ๐‘Ÿ๐‘’๐‘”๐‘–๐‘œ๐‘›)/(๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘Ÿ๐‘’๐‘๐‘ก๐‘Ž๐‘›๐‘”๐‘ข๐‘™๐‘Ž๐‘Ÿ ๐‘ก๐‘–๐‘™๐‘’) Finding area of rectangular region and rectangular tile Rectangular tile Length of tile = 12 cm Breadth of tile = 5 cm Area of tile = Length ร— Breadth = 12 ร— 5 = 60 sq. cm = 60 cm2 Rectangular region Length of region = 100 cm Breadth of region = 144cm Area of region = Length ร— Breadth = 100 ร— 144 = 14400 sq. cm = 14400 cm2 Number of tiles = (๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘Ÿ๐‘’๐‘๐‘ก๐‘Ž๐‘›๐‘”๐‘ข๐‘™๐‘Ž๐‘Ÿ ๐‘Ÿ๐‘’๐‘”๐‘–๐‘œ๐‘›)/(๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘Ÿ๐‘’๐‘๐‘ก๐‘Ž๐‘›๐‘”๐‘ข๐‘™๐‘Ž๐‘Ÿ ๐‘ก๐‘–๐‘™๐‘’๐‘ ) = 14400/60 = 1440/6 = 240 โˆด 240 tiles are required to cover the region Ex 10.3, 12 How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively: (b) 70 cm and 36 cm. Here, Number of tiles = (๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘Ÿ๐‘’๐‘๐‘ก๐‘Ž๐‘›๐‘”๐‘ข๐‘™๐‘Ž๐‘Ÿ ๐‘Ÿ๐‘’๐‘”๐‘–๐‘œ๐‘›)/(๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘Ÿ๐‘’๐‘๐‘ก๐‘Ž๐‘›๐‘”๐‘ข๐‘™๐‘Ž๐‘Ÿ ๐‘ก๐‘–๐‘™๐‘’) From part (a), Area of rectangular tile = 60 cm2 Finding area of rectangular region Rectangular region Length of rectangle = 70 cm Breadth of rectangle = 36 cm Area of rectangle = Length ร— Breadth = 70 ร— 36 = 2520 sq. cm = 2520 ใ€–๐‘๐‘šใ€—^2 Now, Number of tiles = (๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘Ÿ๐‘’๐‘๐‘ก๐‘Ž๐‘›๐‘”๐‘ข๐‘™๐‘Ž๐‘Ÿ ๐‘Ÿ๐‘’๐‘”๐‘–๐‘œ๐‘›)/(๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘Ÿ๐‘’๐‘๐‘ก๐‘Ž๐‘›๐‘”๐‘ข๐‘™๐‘Ž๐‘Ÿ ๐‘ก๐‘–๐‘™๐‘’๐‘ ) = 2520/60 = 252/6 = 42 โˆด 42 tiles are required to cover the given region

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.