Q. 75.0( 4 Votes )

# A tent of height 3.3 m is in the form of a right circular cylinder of diameter 12m and height 2.2m, surmounted by a right circular cone of the same diameter. Find the cost of the canvas of the tent at the rate of Rs. 500 per m^{2}.

Answer :

Given: Height of the tent = 3.3 m

The height of the cylindrical portion = 2.2 m

∴ The height of the conical portion

= Height of the tent – Height of the cylindrical portion

= 3.3 – 2.2

= 1.1m

Given Diameter of the cylinder, d = 12m

CSA of cylindrical portion = 2πrh

= 82.971 m^{2}

Firstly, we have to find the slant height (l) of the conical portion

⇒ l^{2} = h^{2} + r^{2}

⇒ l^{2} = (1.1)^{2} + (6)^{2}

⇒ l^{2} = 1.21 + 36

⇒ l^{2} = 37.21

⇒ l = √37.21

⇒ l = 6.1m

∴ CSA of the conical portion = πrl

= 115.029 m^{2}

So,

Total Surface Area of the tent = Surface area of conical portion + surface Area of cylindrical portion

= 115.029 + 82.971

= 198 m^{2}

∴ Canvas required to make the tent = 198 m^{2}

Cost of 1m^{2} canvas = Rs 500

Cost of 198 m^{2} canvas = Rs 500 × 198

= Rs 99000

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