Question

9Daniel 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..​

Answer 1

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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 2

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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

Thanks !!