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 sq metre of area is painted. How many cans of paint will she need to paint the room ?
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Answers
Answer:
Step-by-step explanation:
Question :-
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 sq metre of area is painted. How many cans of paint will she need to paint the room ?
Given :-
Length of hall - 15m
Breadth of hall - 10m
Height of hall - 7m
Area painted by 1 can = 100m^2
To Find :-
Cans required to paint th room?
Formula Used :-
L.S.A of Cuboid = 2h (l + b)
Area of Rectangle = Length × Breadth
Solution :-
Area of wall to painted.
Area of 4 walls + Area of top
2h(l + b) + (l × b)
2 h (15 + 10) + (15 × 10)
14 × 25 + 150
350 + 150
500m^2
No of cans required = Area of wall / Area of covered by 1 can
500 / 100
=> 5
Hence, No of Cans required to paint the room is 5 cans
.
Answer:
Length= 15m
Breadth= 10m
Height= 7m
Lateral Surface Area + Roof = 2h(l+b)+lb = 2*7(15+10)+15*10
= 14*25 + 150 = 350 + 150 = 500m^2
No. of cans required = 500 / 100 = 5
Therefore, 5 cans are required to paint the room.