Answer :

Given:

A club house dimension,

Length = 7.2 m

Breadth = 5.5 m

Height = 4.2 m

Dimensions of door in the room,

Length = 3 m

Breadth = 1.8 m

Dimensions of each window in the room (there are 2 windows),

Length = 2.25 m

Breadth = 1.8 m

**(A).** Let us measure the area of the floor of the club house.

The floor comprises of 2 dimensions, that is, length and breadth.

It doesn’t have either door or any window.

So, dimensions of floor are

Length = 7.2 m

Breadth = 5.5 m

Then, area of floor is given by

Area of floor = Length × Breadth

⇒ Area of floor = 7.2 × 5.5

⇒ Area of floor = 39.6 m^{2}

Also, let us find total expenditure of cementing the floor at the rate of Rs. 62/m^{2}.

If cost of cementing 1 m^{2} of floor = Rs. 62

Then, cost of cementing 39.6 m^{2} of floor = Rs. 62 × 39.6

⇒ Cost of cementing 39.6 m^{2} of floor = Rs. 2455.2

Hence, area of the floor is 39.6 m^{2} and cost of cementing it is Rs. 2455.2.

**(B).** Let us measure the area of the inner walls excluding the doors and windows.

The inner walls comprise of 4 walls.

Area of first wall = Length × Height

⇒ Area of first wall = 7.2 × 4.2

= 30.24 m^{2}

Area of second wall = Breadth × Height

= 5.5 × 4.2

⇒ Area of second wall = 23.1 m^{2}

Area of third wall = Length × Height

= 30.24 m^{2}

Area of fourth wall = Breadth × Height

= 23.1 m^{2}

Area of inner walls = Area of first wall + Area of second wall + Area of third wall + Area of fourth wall

⇒ Area of inner walls = 30.24 + 23.1 + 30.24 + 23.1

⇒ Area of inner walls = 106.68 m^{2} …(i)

But, the walls also has 1 door and 2 windows.

So, to find the area of the walls excluding the 1 door and 2 windows, just subtract the area of the door and windows from the total area of the walls.

Area of the inner wall (excluding door and windows) = Area of inner wall – (Area of 1 door + Area of 2 windows) …(ii)

Area of 1 door is found out as,

Area of 1 door = Length of door × Breadth of door

⇒ Area of 1 door = 3 × 1.8 [given]

⇒ Area of 1 door = 5.4 m^{2} …(iii)

Area of 1 window is found out as,

Area of 1 window = Length of window × Breadth of window

⇒ Area of 1 window = 2.25 × 1.8 [given]

⇒ Area of 1 window = 4.05 m^{2}

If area of 1 window = 4.05 m^{2}

Then, area of 2 windows = 2 × 4.05

⇒ Area of 2 windows = 8.1 m^{2} …(iv)

Substituting equations (i), (iii) and (iv) in equation (ii), we get

Area of inner walls (excluding door and windows) = 106.68 – (5.4 + 8.1)

⇒ Area of inner walls (excluding door and windows) = 106.68 – 13.5

⇒ Area of inner walls (excluding door and windows) = 93.18 m^{2}

Hence, the area of inner walls excluding door and windows is 93.18 m^{2}.

**(C).** To find the area of the ceiling of the room, we need to know the length and breadth of the ceiling.

Length = 7.2 m

Breadth = 5.5 m

So, area of the ceiling is given by

Area = Length of the ceiling × Breadth of the ceiling

⇒ Area = 7.2 × 5.5

⇒ Area = 39.6 m^{2}

Note that, area of the ceiling is equal to the area of the floor.

Hence, area of the ceiling is 39.6 m^{2}.

**(D).** Now, we need to find the expenditure in whitewashing inner walls and ceiling without door and windows.

Let us recall that,

Cost of whitewashing 1 m^{2} = Rs. 12

We have,

Area of inner walls (excluding door and windows) = 93.18 m^{2}

Area of ceiling = 39.6 m^{2}

Total area for whitewashing = Area of inner walls (excluding door and windows) + Area of ceiling

⇒ Total area for whitewashing = 93.18 + 39.6

⇒ Total area for whitewashing = 132.78 m^{2}

If cost of whitewashing 1 m^{2} = Rs. 12

Then, cost of whitewashing 132.78 m^{2} = Rs. 12 × 132.78

⇒ Cost of whitewashing 132.78 m^{2} = Rs. 1593.36

Hence, total amount for whitewashing inner walls and ceiling except door and windows is Rs. 1593.36.

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