Question

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 2.1m 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 500 per m². (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 cylinder and a cone.

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.

Answer 2

$\huge\mathcal{\fcolorbox{gold}{black}{\red{❥ᴀɴsᴡᴇʀ࿐}}}$

It is known that a tent is a combination of cylinder and a cone.

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.

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Mark upper one Brainliest $\huge\color{red}♡$