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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

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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo