the radius of cylindrical cistern to store water is 35 CM & it's height is 80 CM find the cost of painting its curved surface at rate rs 2 per 100 cm square
Answers
Answer:
Q1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It the top. Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required .for snaking the box.
(ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs. 20.
Sol: (i) Here, l = 1.5 m, b = 1.25 m
∵ It is open from the top.
∵ Its surface area = [Lateral surface area] + [Base area]
= [2(l + b)h] + [l × b]
= [2(1.50 + 1.25)0.65 m2] + [1.50 × 1.25 m2]
= [2 × 2.75 × 0.65 m2] + [1.875 m2]
= 3.575 m2 + 1.875 m2 = 5.45 m2
∵ The total surface area of the box = 5.45 m2
∴ Area of the sheet required for making the box = 5.45 m2
(ii) Rate of sheet = Rs. 20 per m2
Cost of 5.45 m2 = Rs. 20 × 5.45
⇒ Cost of the required sheet = Rs.109
Q2. 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 off 7.50 per m2.
Sol: Length of the room (l) = 5m
Breadth of the room (b) = 4 m
Height of the room (h) = 3 m
The room is like a cuboid whose four walls (lateral surface) and ceiling are to be white washed.
∴ Area for white washing = [Lateal surface area] + [Area of the ceiling]
= [2(1 + b)h] + [1 × b]
= [2(5 + 4) } 3 m2] + [5 × 4 m2]
= [54m2] + [20m2] = 74m2
Cost of white washing:
Cost of white washing for 1 m2 = Rs. 7.50
∴ Cost of white washing for 74 m2 = Rs. 7.50 × 74
The required cost of white washing is Z555.
Q3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of 10 per m2 is Rs. 15000, find the height of the hall.
Sol: Note: Area of four walls = Lateral surface area.
A rectangular hall means a cubiod.
Let the length and breadth of the hall be ‘l’ and ‘b’ respectively.
∵[Perimeter of the floor] = 2(l + b)
⇒ [2(l + b)] = 250 m.
∵Area of four walls = lateral surface area
⇒ [2(l + b)] × h [where ’h’ is the height of hall.]
Cost of painting the four walls = Rs. 10 × 250 h = Rs. 2500 h
⇒ Rs. 2500 h = Rs. 15000
Thus, the required height of the hall 6 m
Q4. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Sol: Total area that can be painted = 9.375 m2, since a brick is like a cuboid
∴ Total surface area of a brick = 2[lb + bh + hl]
= 2[(22.5 × 10) + (10 × 7.5) + (7.5 × 22.5)] cm2
= 2[(225) + (75) + (168.75)] cm2
Thus, the required number of bricks = 100
Q5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
Sol: For the cubical box
∵Edge of the cubical box = 10 cm
∴ Lateral surface area = 4a2
= 4 × 102 cm2
= 4 × 100 cm2
= 400 cm2
Total surface area = 6a2
= 6 × 102 = 6 × 100 cm2
= 600 cm2
∵For the cuboidal box, l = 12.5 cm, b = 10 cm, h = 8 cm
∴ Lateral surface area = 2[(l + b)] × h
= 2[12.5 + 10] × 8 cm2
= 2[22.5 × 8] cm2
= 360 cm2
Total surface area = 2[lb + bh + hl]
= 2[(12.5 × 10) + (10 × 8) + (8 × 12.5)] cm2
= 2[125 + 80 + 100] cm2
= 2[305] cm2 = 610 cm2
Obviously,
(i) ∵ 400 cm2 > 360 cm2 and 400 – 360 = 40
∴ The cubical box, has greater lateral surface area by 40 m2.
(ii) ∵ 610 cm2 > 600 cm2 and 610 – 600 = 10
∴The cuboidal box has greater total surface area by 10 m2.
Q6. 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?
Sol: The herbarium is like a cuboid
Here, l = 30 cm, b = 25 cm, h = 25 cm
(i) ∵ Area of a cuboid = 2[lb + bh + hl]
∴ Surface area of the herbarium (glass) = 2[(30 × 25) + (25 × 25) + (25 × 30)] cm2
= 2[750 + 625 + 750] cm2
= 2[2125] cm2
= 4250 cm2
Thus, the required area of glass is 4250 cm2.
(ii) Total length of 12 edges = 4l + 4b + 4h
= 4(1 + b + h) = 4(30 + 25 + 25) cm
= 4 × 80 cm = 320 cm
Thus, length of tape needed = 320 cm