Dawood is painting the walls and ceiling of a cuboidal hall with length,
breadth and height of 15m, 10m and 7m respectively. From each can of paint
100m2
of area is painted. How many cans of paint will be needed to paint the
room.
Answers
Answer:
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.
Answer:
5 cans
Step-by-step explanation:
Mensuration :
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...