Question

Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height

of 15 m, 5 m and 7 m respectively. From each can of paint 100 m2 of area is painted.

How many cans of paint will she need to paint the room?​

Answer 1

Answer:

L=15m ; B=5m; H=7m

Area painted by 1 can =100m^2

No. of cans required=Area of the hall/ Area painted by 1 can .

Area of the hall = Total Surface Area of hall-Area of bottom

Area of the hall =2(lb+bh+bl)

2(15*5+5*7+7*15)

2(75+35+105)

2*215=430m^2

Area of bottom=l*b=15*5=75m^2

Area of the hall =430-75m^2

=355m^2

Total numbers of cans =Area of the hall / area painted by one can

= 355m^2/100=3.55

3.55 cans would be required to paint the room.

But I think that your question is wrong ..it should be B=10 m and if the B=10m then

Area of hall =2(15*10+10*7+7*15)=2(150+70+105)=2*325=650m^2

Area of bottom=l*b=15*10=150m^2

Area of hall =650-150m^2=500m^2

Total no.of cans =500m^2/100=5

So, 5 cans would be required to paint the room.