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¨πππ ππ πππ π―πππ)/(π¨πππ πππππππ ππ π πππ)
Finding Area of Hall Painted
Given that Daniel paints wall and ceiling of hall
He doesnβt 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(ππ+πβ+βπ)
= 2(15Γ10+10Γ7+15Γ7)
= 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¨πππ ππ πππ π―πππ)/(π¨πππ πππππππ ππ π πππ)
= 500/100
= 5
So, 5 cans are required

Made by

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.

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