metal piece is 28 cm long and has an external radius of 2.1 CM if its thickness is 0.7 CM then find the whole surface area of the pipe x
Answers
Answer:
PSP~♡
Step-by-step explanation:
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.
Answer:
Solution: Total surface area=CSA of outside cylinder+ CSA of inside cylinder+2*area of ring. `=2piRh+2pirh+2pi(r^2-r^2)` `=2pi(25/2h+23/2h+(25/2+23/2)(25/2-23 ..
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