"Question 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 m^2 of area is painted. How many cans of paint will she need to paint the room?
Class 8 Mensuration Page 186"
Answers
Mensuration:
Mensuration is the branch of mathematics which concerns itself with the measurement of Lengths, areas & volume of different geometrical shapes or figures.
Surface area:
The surface area of a solid is the sum of the areas of the plane and curved faces of the solid.
It is measured in square units such as square centimetre (cm²) and square metre (m²).
Surface area of cuboid is the sum of the surface areas of its six rectangular faces.
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.
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Hope this will help you...
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