Question

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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.)​

Answer 1

Answer:

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

So, Rs. 22000 will be the total cost of the canvas

Answer 2

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Area of the canvas = Curved Area of Cone + Curved Surface Area of Cylinder

Curved Surface Area of Cone :-

=> Diameter of cone = Diameter of cylinder = 4cm

=> Radius = Diameter / 2 = 4/2 = 2cm

=> Slant Height = l = 2.8 m

Curved Surface Area of cone :-

=> πrl

=> 22/7 × 2 × 2.8

=> 22 × 2 × 0.4

=> 17.6m^2

Curved Surface Area of Cylinder :-

=> Diameter of Cylinder = 4cm

=> Radius = Diameter / 2 = 4/2 = 2cm

=> Height = 2.1 m

Curved Surface Area of Cylinder :-

=> 2πrh

=> 2 × 22/7 × 2 × 2.1

=> 2 × 22 × 2 × 21 / 7 × 10

=> 132/5

=> 26.4 m^2

Area of the Canvas :-

=> CSA of Cone + CSA of Cylinder

=> 17.6 × 26.4

=> 44 m^2

Cost of canvas for tent for 1 m^2 = ₹500

Cost of canvas for tent for 44 m^2 = ₹500 × 44

=> 22,000

Hence , total cost is ₹22,000

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