Ex 10.3, 12 - Chapter 10 Class 6 Mensuration - Part 5

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Ex 10.3, 12 - Chapter 10 Class 6 Mensuration - Part 6

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Ex 10.3, 12 - Chapter 10 Class 6 Mensuration - Part 7

  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: (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 10 years. He provides courses for Maths and Science at Teachoo.