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 ?
Please give good answers only..
Answers
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
.
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