A hall of length 20ft, breadth 10ft and height 7ft is to be painted along walls. How many cans of paint
are required to paint the walls if from each can 30ft2
of area is painted? What will be the cost of
painting if cost of each can is Rs. 75?
Answers
Answer:
Surface Area of Cuboid is =
2(lb + lh + bh)
By surface area of a cuboid we mean the total surface area.
The sum of the areas of 4 vertical faces of a cuboid is called its lateral surface area.
The lateral surface area or the area of the four walls of the cuboid .
Area of the four walls = 2 (l +b) h
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Solution:
Given;
Length of wall (l)= 15 m, Breadth of wall (b)= 10 m Height of wall (h) = 7 m
Total area to be painted= area of 4 walls + area of ceiling
= 2(l+b)h + lb
= 2(15+10)7 + 15×10
= 2×25×7+ 150
=50× 7+150= 350+ 150=500
Total area to be painted=500m²
Given 100m² area can be painted from each can.
Number of cans Required=
Area of hall/ area of 1 can
= 500/100= 5
Hence, 5 cans are required to paint the room.
==========================================================
Hope this will help you...
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cans
You have to use the open cuboid formula
=2h(l+b) +lb
=2*7(15+10)+ (15)(10)
=14*25+150
=490
No. of cans required = 490/100
= 4.90 cans
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