A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)
Answers
Given:
Height (h) of the cylindrical part = 2.1 m
Diameter of the cylindrical part = 4 m
Radius of the cylindrical part = 2 m
Slant height (l) of conical part = 2.8 m
Area of canvas used = CSA of conical part + CSA of cylindrical part
=πrl+2πrh
=π×2×2.8+2π×2×2.1
=2π[2.8+4.2]
=2× 7 x 22/7 x 7
=44m^2
Cost of 1 m^2 canvas=500 Rs.
Cost of 44m^2 canvas= 44 x 500=Rs. 22000
Therefore, it will cost 22000 rupees for making such a tent.
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Solution :-
It is known that a tent is a combination of cylinder and a cone.
[ Refer to the attachment ]
From the question we know that
- Diameter = 4 m
- Slant height of the cone (l) = 2.8 m
- Radius of the cone (r) = Radius of cylinder = 4/2 = 2 m
- Height of the cylinder (h) = 2.1 m
So, the required surface area of tent = surface area of cone + surface area of cylinder
=> πrl+2πrh
=> πr(l+2h)
=> (22/7)×2(2.8+2×2.1)
=> (44/7)(2.8+4.2)
=> (44/7)×7 = 44 m² .
∴ The cost of the canvas of the tent at the rate of ₹500 per m² will be.
=> Surface area × cost per m²
=> 44×500 = ₹22000
So, Rs. 22000 will be the total cost of the canvas.
