Ex 9.2

Chapter 9 Class 8 Mensuration
Serial order wise

### Transcript

Ex 9.2, 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 = π = 7 m Also, Area painted by 1 can = πππ π^π 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 = π(ππ+ππ+π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