1).A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine: i) The area of the sheet required for making the box. ii) The cost of sheet for it, if a sheet measuring 1 m² costd Rs 20?
Solution :
i) We have,
l = Length of plastic box = 1.5 m
b = Breadth of plastic box = 1.25 m
h = height of plastic box = 65cm = 65/100 = 0.65 m
then, total surface area of the plastic box, when it is open at top = lb + 2 ( bh + lh )
= [ 1.5 × 1.25 + 2 ( 1.25 × 0.65 + 0.65 × 1.5)] m²
= [ 1.875 + 2 ( 0.8125 + 0. 975) ] m²
= [ 1.875 + 3.575] m² = 5.45 m²
Hence, the area of the sheet required for making the box is 5.45 m²
ii) ∵ Cost of 1 m² sheet = Rs. 20
∴ Cost of 5.45 m² sheet = 5.45 × 20 = Rs 109
Hence, the cost of sheet required for making the box is Rs 109.
2) The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs7.50 m²?
Solution : Given,
Length of the room, l = 5 m
Breadth of the room, b = 4 m
and height, h = 3 m
Now,
Area of the 4 wall = 2 ( l + b ) × h
= 2 ( 5 + 4 ) × 3 = 18 × 3 = 54 m²
and area of the ceiling = l × b = 5 × 4 = 20 m²
Then,
total area ( 4 wall + ceiling ) = 54 + 20 = 74 m²
∵ Cost of white washing in 1 m² = Rs 7.50
∴ Cost of washing in 74 m² = 74 × 7.50 = Rs.555
Hence, the cost of white washing the walls and the ceiling is Rs. 555
3) The floor of a rectangular hall has a perimeter 250m. If the cost of painting the four walls at the rate of 10 per m² is Rs 15000. Find the height of the hall?
Solution :
Let the dimensions of the floor of the hall be l meter × w metre × h meter
Area of four walls of the hall = 2 (l × h) + 2(w × h) = 2 l h + 2 w h = {2(l + w)} × h
Now 2 (l + w ) = perimeter of the floor of the hall = 250 m
Area of four walls of the hall = 250 h m²
Cost of painting of four walls of the hall @ of Rs 10 m² = Rs 15,000
Area of the four walls = Rs 15,000 /Rs 10 per m² = 1500 m²
According to above
250 h = 1500
h = 1500/250 = 6 m.
Hence, the height of the hall is 6 m.
4). The paint in a certain container is sufficient to paint on area equal to 9.375 m². How many bricks of dimension 22.5cm × 10cm × 7.5 cm can be painted out of this container?
Solution :
Given,
Area of the container = 9.375 m²
= 9.375 × 100 × 100 cm² = 93,750 cm²
Now,
Surface area of one bricks = Total surface area of the cuboid , where length, l = 22.5 cm, breadth, b = 10 cm and height, h = 7.5 cm
∵ Surface area of one bricks =
2 ( lb + bh + lh) = 2 ( 22.5 × 10 + 10 × 7.5 + 7.5 × 22) = 2 × 468.75 = 937.5 cm²
Hence, the number of bricks =
= ( Surface area of container)/( Surface area of one bricks)
= 93750/937.5 = 100
Hence, the number of bricks = 100
5). A cubical box has each edge 10cm and another cuboidal box is 12.5 cm long, 10cm wide and 8 cm high. i) which box has the smaller total surface area and by how much ii) which box has the greater lateral surface area and by how much?
Solution :
Given,
For cubical box, edge of the cube, a = 10 cm
For cubodial box, length, l = 12.5 cm, breadth, b = 10 cm and height, h = 8 cm
i) Let S₁ and S₂ be the total surface area of the cubical and cuboidal box respectively.
∴ S₁ = 6 a² = 6 × 10² = 600 cm²
and S₂ = 2 ( lb + bh + lh )
or, S₂ = 2 ( 12.5 × 10 + 10 × 8 + 8 × 12.5)
= 2 ( 125 + 80 + 100) = 610 cm²
∴ S₂ - S₁ = 610 - 600 = 10 cm²
Hence, the cuboidal box has greater total surface area and is greater by 10 cm².
ii) Let L₁ and L₂ be the lateral surface areas of the cubical and cubodial boxes respectively.
Then,
L₁ = 4 a² = 4 × 10² = 400 cm²
L₂ = 2 ( l + b ) × h = 2 ( 12.5 + 10) × 8
= 360 cm²
∴ L₁ - L₂ = ( 400 - 360 ) cm² = 40 cm²
Hence, the cubical box has larger lateral surface area and is greater by 40 cm².
6). A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high. i) what is the area of the glass ii) How much of tape is needed for all the 12 edges?
Solution :
Given,
Length, l = 30 cm, breadth, b = 25cm and height, h = 25cm
Now, i)
Area of the glass = Total surface area of cuboid of length, l, breadth, b and height, h
∴ Area of the cuboid = 2( lb + bh + lh)
= 2( 30 × 25 + 25 × 25 + 30 × 25)
= 2( 750 + 625 + 750) = 4250 cm²
ii) Length of the tape = Length of all 12 edges = 4 ( l + b + h)
= 4( 30 + 25 + 25) = 320 cm.
Hence, the area of the glass is 4250 cm² and length of the tape is 320 cm
7). Shanti Sweets stall was placing an order for making card board boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25cm × 20cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. 5% of the total surface area is required extra, for all the overlaps. If the cost of cardboard is Rs 4 for 1000 cm² find the cost of cardboard required for supplying 250 boxes of each kind?
Solution :
Let S₁ and S₂ are the surface areas of two boxes.
For S₁ box, l₁ = 25 cm, b₁ = 20 cm and h₁ = 5 cm
For S₂ box, l₂ = 15 cm, b₂ = 12 cm and h₂ = 5 cm
Now, S₁ = 2 ( l₁b₁ + b₁h₁ + l₁h₁ )
= 2( 25. 20 + 20.5 + 25.5) = 1450 cm²
S₂ = 2( l₂b₂ + b₂h₂ + l₂h₂)
= 2( 15 × 12 + 12 × 5 + 15 × 5) = 630 cm²
∴ S₁ + S₂ = 1450 + 630 = 2080 cm²
∵ Area of overlaps = 5% of ( S₁ + S₂ )
= 5/100 × 2080 = 104 cm²
Total surface area of two boxex( one box of each kind) = 2080 + 104 = 2184 cm²
∴ Surface area of 250 boxes = 250 × 2184 = 546000 cm²
Now, Cost of cardboard for 1000 cm²= Rs 4
∴ Cost of cardboard for 1 cm² = 4/1000
∴ Cost of cardboard for 250 boxes =
4/1000 × 546000 = Rs 2184
Hence, the cost of cardboard for 250 boxes of each kind = Rs 2184
8). Parveen wanted to make a temporary shelter for his car by making a box - like structure with tarpaulin that covers all the four sides and the top of the car ( with the front face as a flap which can be rolled up) . Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m with base dimensions 4 m × 3 m ?
Solution : Given,
Length of the tarpaulin, l = 4 m
Breadth of the tarpaulin, b = 3 m
Height of the tarpaulin, h = 2.5 m
Now,
Lateral surface area of tarpaulin = Lateral surface area of cuboid = 2 ( l + b ) × h
= 2 ( 4 + 3) × 2.5 = 35 m²
∵ Area of roof = Area of rectangle = l × b
= 4 × 3 m² = 12 m²
Then,
Area of tarpaulin = Lateral surface area of tarpaulin + Area of roof
= 35 m² + 12 m² = 47 m²
Hence, the area of the tarpaulin is 47 m².
