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Ex 11.3, 5 - Daniel is painting walls and ceiling of a cuboidal hall

Ex 11.3, 5 - Chapter 11 Class 8 Mensuration - Part 2
Ex 11.3, 5 - Chapter 11 Class 8 Mensuration - Part 3
Ex 11.3, 5 - Chapter 11 Class 8 Mensuration - Part 4

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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 = 𝑙 = 15 m Breadth of hall = 𝑏 = 10 m & Height of hall = β„Ž = 7 m Also, Area painted by 1 can = 100 π‘š^2 Now, Number of cans required = (𝑨𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝑯𝒂𝒍𝒍)/(𝑨𝒓𝒆𝒂 π’‘π’‚π’Šπ’π’•π’†π’… π’ƒπ’š 𝟏 𝒄𝒂𝒏) Finding Area of Hall Painted Given that Daniel paints wall and ceiling of hall He doesn’t paint bottom So, Area Painted = Total surface Area of Hall – Area of bottom Total surface area of hall Area = Total surface area of cuboid = 2(𝑙𝑏+π‘β„Ž+β„Žπ‘™) = 2(15Γ—10+10Γ—7+15Γ—7) = 2(150+70+105) = 2(325) = 650 m2 Bottom area of hall Area = Length Γ— Breadth = 15 Γ— 10 = 150 m2 Now, Area painted = Total surface area βˆ’ Bottom area = 650 – 150 = 500 m2 Thus , Number of cans required = (𝑨𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝑯𝒂𝒍𝒍)/(𝑨𝒓𝒆𝒂 π’‘π’‚π’Šπ’π’•π’†π’… π’ƒπ’š 𝟏 𝒄𝒂𝒏) = 500/100 = 5 So, 5 cans are required

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.