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 ?

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Answer 1

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

.

Hope it helpful.. ✌️

@Somya2563

#Be Brainly

Answer 2

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