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 cane of paint 100 2 of area is painted. How many canes of

paint will be required to paint the entire room.​

Answer 1

Given :-

Length of the wall = 15 m

Breadth of the wall = 10 m

Height of the wall = 7 m

To Find :-

Canes of  paint will be required to paint the entire room.​

Solution :-

We know that,

• l = Length
• b = Breadth
• h = Height

By the formula,

$\underline{\boxed{\sf TSA \ of \ cuboid=lb+2(bh+hl ) }}$

Given that,

Length (l) = 15 m

Breadth (b) = 10 m

Height (h) = 7 m

Substituting their values,

= 15 × 10 + 2(10 × 7 + 7 × 15)

= 150 + 2(70 + 105)

= 150 + 350

= 500

Required number of cans =  Area of hall/Area of one can

Substituting them,

= 500/100 = 5

Therefore, 5 cans are required to paint the room.

Answer 2

$\huge \purple A \purple N \purple s \purple W \purple e \purple \R:-$

GIVEN ;

length of wall = l = 15 m

breadth of wall = b = 10m

height of wall = h = 7m

Area painted by 1 can = 100m^2

TO FIND ;

How many cans required to paint entire room .

ANSWER ;

Number of cans required = Area of the hall / Area painted by 1 can

finding area of hall painted:-

Daniel paints wall and ceiling of the hall

He doesn't paint the bottom .

Area painted = TSA of hall - Area of bottom

TSA OF HALL :-

Area = TSA of cuboid

= 2 ( lb + bh + hl )

= 2 ( 15×10 + 10×7 + 7×15 )

= 2 ( 150 + 70 + 105 )

= 2 × 325

= 650 m^2

BOTTOM AREA OF HALL :-

Area = l × b

= 15 × 10

= 150 m^2

NOW :-

Area painted = TSA - bottom area

= 650 - 150

= 500 m^2

Thus ,

Number of cans required = Area of hall / Area painted by 1 can

= 500 / 100

= 5

THEREFORE :- 5 cans are required