Question

Avantika is painting the walls and ceiling of a cuboidal greenhouse with length, breadth and height of 15m, 9m and 7m respectively. From each can of paint 100m^2 of area is painted. How man
cans of paint will she need to paint the room?

Answer 1

Answer:

$\bigstar{\bold{Number\:of\:cans\approx 5\:cans}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Length of the green house = 15 m
• Breadth of the green house = 9 m
• Height of the green house = 7 m
• Area that can be painted with one can of paint =  100 m²

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Number of cans of paint needed to paint the room

$\Large{\underline{\underline{\bf{Solution:}}}}$

➤ First we have to find the area of the room to be painted.

➤ Area to be painted = Ceiling + 4 walls

➤ Area to be painted = lb + 2h (l + b)

➤ Substitute the given data,

    Area to be painted = 15 × 9 + 2 × 7 (15 + 9)

    Area to be painted = 135 + 14 × 24

    Area to be painted = 135 + 336

    Area to be painted = 471 m²

➤ Hence the area to be painted is 471 m².

➤ Now we have to find the number of cans required.

➤ By given we know that 100 m² area can be painted with each can

➤ Hence,

    Number of cans = Area to be painted/100

➤ Substitute the data,

    Number of cans = 471/100

    Number of cans = 4.71

➤ Hence approximately 5 cans will be required to paint the green house.

$\boxed{\bold{Number\:of\:cans\approx 5\:cans}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The total surface area of a cuboid is given by

   Total surface area of a cuboid = 2 (lh + lb + bh)

→ The lateral surface area of a cuboid is given by

   Lateral surface area of a cuboid = 2h (l + b)

→ The Volume of a cuboid is given by

   Volume of a cuboid = l × b × h

   where l = length of the cuboid

              b = breadth of the cuboid

              h = height of the cuboid

Answer 2

Answer :-

» Number of cans of paint required ≈ 5

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★Concept :-

• Here the concept of Surface area and mensuration is used. According to this, if we need to find the area of any given figure or thing, we can directly apply the formula using the given measures of its dimensions.

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★Solution :-

Given,

• Length of the side of greenhouse = l

= 15 m

• Breadth of the side of greenhouse = b

= 9 m

• Height of the side of greenhouse = h

= 7 m

• Minimum area that can be painted using one can = 100 m²

________________________________

» First of all, we need to find out the total area to be painted. This is given by :-

✒ Area to be painted = Area of ceiling + Area of 4 walls joined end to end with each other

Here we aren't assuming the floor that it is painted because normally we don't usually paint the floor or a house or room.

✒ Area to be painted = (l × b) + 2h(l + b)

✒ Area to be painted = (15 × 9) + 2(7)×(15+ 9)

✒ Area to be painted = 135 + (14 × 24)

✒ Area to be painted = 135 + 336

✒ Area to be painted = 471 m²

✳ Here we get, the area from greenhouse, that is to be painted =

= 471 m²

________________________________

• Now we need to find out the number of cans required to paint the area.

We know that, if we divide the total area to be painted by the minimum area painted by a single can, we can get the number of cans.

This is because, the requirement of cans directly depend upon the area to be painted. Then,

▶ Number of cans =

= Area to be painted / (Minimum area painted by a single can)

▶Number of cans = {471 m²} / {100 m²}

▶Number of cans = 4.71

Since, the quantity of cans cannot be in fraction, we have to round off the result. So,

▶ Number of cans = 4.71 ≈ 5

★ Hence, the number of cans required to paint the greenhouse

= 5

___________________________

★More to know :-

• Volume of Cuboid = Length × Breadth × Height

• Volume of Cube = (Side)³

• Volume of Right Circular Cylinder = πr²h

where, r is the radius and h is height.

• Volume of Right Circular Cone = ⅓(πr²h)

where, r is the radius and h is the height

• Area of Rectangular Plane = Length × Breadth

• Area of Square Plane = (Side)²

• Area of Circular Plane/Base = πr²

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