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 4m respectively, and the slant height of the top is 2.8m find the area of the canvas of the tent at the rate of Rs. 500 per square m.

Answer 1

${\underline{\sf{Given}}}$

• Slant height of cone = 2.8 m
• Height and diameter of the cylindrical part are 2.1 m and 4 m
• Cost for the canvas of the tent for 1m² = ₹500 per m²

${\underline{\sf{To\:Find}}}$

• Area of the Canvas required for making the tent.
• The the cost of the Canvas of the tent at the rate of ₹500 per m².

${\underline{\sf{Solution}}}$

We have , Diameter of cone = 4m

⇒ Radius of cone = 2m

Slant Height of cone = 2.8 m

We know that

Curved surface area of cone

$\sf = \pi \: rl$

$\sf = \dfrac{22}{7} \times 2 \times 2.8$

$\sf = 22 \times 2 \times 0.4$

$\sf = 17.6m {}^{2}$

Given :

Diameter of cylinder = 4 m

⇒ Radius of cylinder = 2 m

and height of cylinder = 2.1 m

Curved surface area of cylinder

$\sf = 2\pi \: rh$

$\sf = 2 \times\dfrac{22}{7} \times 2 \times 2.1$

$\sf = \dfrac{22 \times 2 \times 2 \times 21}{7 \times 10}$

$\sf = \dfrac{132}{5}$

$\sf = 26.4m {}^{2}$

Thus ,

Area of Canvas

= Curved surface area of the cylinder+

curved surface area of the cone

= 17.6+26.4

= 44m²

Now ,

Cost for the canvas of the tent for 1m² = ₹500 per m²

⇒Cost for the canvas of the tent for 44m²

= ₹500× 44 per m²

= ₹22,00 m²

$\rule{200}2$

${\red{\boxed{\large{\bold{Note}}}}}$

The base of the tent will not be covered with the Canvas.

Answer 2

Given :

• A tent is in the shape of a cylinder surmounted by a conical top .
• The height and diameter of the cylindrical part are 2.1 m and 4m respectively.
• The slant height of the top is 2.8 m.

To find :

• Area of the Canvas required for making the tent.
• The cost of the Canvas of the tent at the rate of ₹500 per m².

Solution :

• Height(h) of the cylindrical part = 2.1 m
• Diameter of the cylindrical part = 4 m.
• Slant height(l) of the conical part = 2.8 m.

Therefore,

• Radius(r) of the cylindrical part = 4/2 = 2 m.

✪ CSA of the cone ,

$\implies\sf{\pi\:r\:l}$

$\implies\sf{(\dfrac{22}{7}\times\:2\times\:2.8)\:m^2}$

$\implies\sf{17.6\:m^2}$

✪ CSA of cylinder ,

$\implies\sf{2\pi\:r\:h}$

$\implies\sf{(2\times\dfrac{22}{7}\times\:2\times\:2.1)\:m^2}$

$\implies\sf{26.4\:m^2}$

Then,

Area of the canvas used ,

= CSA of cone + CSA of cylinder

= (17.6 + 26.4) m²

= 44 m²

The cost of the Canvas of the tent at the rate of ₹500 per m² ,

= ₹ ( 500× 44)

= ₹ 22,000

Therefore, the area of the Canvas required for making the tent is 44 m² and the cost of the canvas of the tent at the rate of ₹ 500 per m² is ₹ 22,000.