Surface Area of Cuboid

Chapter 11 Class 8 Mensuration
Concept wise

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### 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 = π = 15 m Breadth of hall = π = 10 m & Height of hall = β = 7 m Also, Area painted by 1 can = 100 π^2 Now, Number of cans required = (π¨πππ ππ πππ π―πππ)/(π¨πππ πππππππ ππ π πππ) Finding Area of Hall Painted Given that Daniel paints wall and ceiling of hall He doesnβt paint bottom So, Area Painted = Total surface Area of Hall β Area of bottom Total surface area of hall Area = Total surface area of cuboid = 2(ππ+πβ+βπ) = 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 β Bottom area = 650 β 150 = 500 m2 Thus , Number of cans required = (π¨πππ ππ πππ π―πππ)/(π¨πππ πππππππ ππ π πππ) = 500/100 = 5 So, 5 cans are required

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.