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  1. Chapter 11 Class 8 Mensuration
  2. Serial order wise

Transcript

Ex 11.3, 5 Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m2 of area is painted. How many cans of paint will she need to paint the room? Given, Length of hall = ๐‘™ = 15 m Breadth of hall = ๐‘ = 10 m & Height of hall = โ„Ž = 7 m Also, Area painted by 1 can = 100 ๐‘š^2 Now, Number of cans required = (๐‘จ๐’“๐’†๐’‚ ๐’๐’‡ ๐’•๐’‰๐’† ๐‘ฏ๐’‚๐’๐’)/(๐‘จ๐’“๐’†๐’‚ ๐’‘๐’‚๐’Š๐’๐’•๐’†๐’… ๐’ƒ๐’š ๐Ÿ ๐’„๐’‚๐’) Finding Area of Hall Painted Given that Daniel paints wall and ceiling of hall He doesnโ€™t paint bottom So, Area Painted = Total surface Area of Hall โ€“ Area of bottom Total surface area of hall Area = Total surface area of cuboid = 2(๐‘™๐‘+๐‘โ„Ž+โ„Ž๐‘™) = 2(15ร—10+10ร—7+15ร—7) = 2(150+70+105) = 2(325) = 650 m2 Bottom area of hall Area = Length ร— Breadth = 15 ร— 10 = 150 m2 Now, Area painted = Total surface area โˆ’ Bottom area = 650 โ€“ 150 = 500 m2 Thus , Number of cans required = (๐‘จ๐’“๐’†๐’‚ ๐’๐’‡ ๐’•๐’‰๐’† ๐‘ฏ๐’‚๐’๐’)/(๐‘จ๐’“๐’†๐’‚ ๐’‘๐’‚๐’Š๐’๐’•๐’†๐’… ๐’ƒ๐’š ๐Ÿ ๐’„๐’‚๐’) = 500/100 = 5 So, 5 cans are required

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.