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 m2
Also, let us find total expenditure of cementing the floor at the rate of Rs. 62/m2.
If cost of cementing 1 m2 of floor = Rs. 62
Then, cost of cementing 39.6 m2 of floor = Rs. 62 × 39.6
⇒ Cost of cementing 39.6 m2 of floor = Rs. 2455.2
Hence, area of the floor is 39.6 m2 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 m2
Area of second wall = Breadth × Height
= 5.5 × 4.2
⇒ Area of second wall = 23.1 m2
Area of third wall = Length × Height
= 30.24 m2
Area of fourth wall = Breadth × Height
= 23.1 m2
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 m2 …(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 m2 …(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 m2
If area of 1 window = 4.05 m2
Then, area of 2 windows = 2 × 4.05
⇒ Area of 2 windows = 8.1 m2 …(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 m2
Hence, the area of inner walls excluding door and windows is 93.18 m2.
(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 m2
Note that, area of the ceiling is equal to the area of the floor.
Hence, area of the ceiling is 39.6 m2.
(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 m2 = Rs. 12
We have,
Area of inner walls (excluding door and windows) = 93.18 m2
Area of ceiling = 39.6 m2
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 m2
If cost of whitewashing 1 m2 = Rs. 12
Then, cost of whitewashing 132.78 m2 = Rs. 12 × 132.78
⇒ Cost of whitewashing 132.78 m2 = Rs. 1593.36
Hence, total amount for whitewashing inner walls and ceiling except door and windows is Rs. 1593.36.
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