Question

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 m^2. (Note that the base of the tent will not be covered with canvas.)​

Answer 1

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It is known that a tent is a combination of a cone and a cylinder as shown in the figure.

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

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

$= \pi \: rl + 2\pi \: rh$

$= \pi \times 2 \times 2.8 + 2\pi \times 2 \times 2.1$

$= 2\pi(2.8 + 4.2)$

$= 44 \: {m}^{2}$

Cost of 1 m2 canvas =500 rupees

Cost of 44 m2 canvas = 44 × 500 = 22000

Therefore, it will cost 22000 rupees for making such a tent.

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