7. 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
Step-by-step explanation:
ANSWER
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
22
×7
=44m
2
Cost of 1 m
2
canvas =500 rupees
Cost of 44 m
2
canvas =44×500=22000 rupees.
Therefore, it will cost 22000 rupees for making such a tent.
From the question, we know that
The diameter = D = 4 m
l = 2.8 m (slant height)
The radius of the cylinder is equal to the radius of the cylinder
So, r = 4/2 = 2 m
Also, we know the height of the cylinder (h) is 2.1 m
So, the required surface area of the given tent = surface area of the cone (the top) + surface area of the cylinder(the base)
= πrl + 2πrh
= πr (l+2h)
Now, substituting the values and solving it we get the value as 44 m^2
∴ The cost of the canvas at the rate of Rs. 500 per m^2 for the tent will be
= Surface area × cost/ m^2
= 44 × 500