Answer :

We have,

Height of the cylinderical part = 2.1 m

Diameter of the cylinderal part = 4 m

Radius of the cylinerical part = 2 m

Slant height of the conical part = 2.8 m

Thus, Total canvas used = Curved surface area of cylinder + Curved surface area of cone

\( = \ 2 \pi rh \ + \ \pi rl \ = \ \pi r(2h \ + \ l) \)

\(= \ \frac{22}{7} \ × \ 2 \ × \ (2 \ × \ 2.1 \ + \ 2.8) \ \)

\( = \ 44 \) m^{2}

Now, cost of 1 m^{2} the canvas for the tent = Rs 500

So, cost of 44 m^{2} the canvas for the tent = Rs 44 × 500 = Rs 22000

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- NCERT solutions for class 10 science chapter 2 Acids, Bases and Salts
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