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A tent is made in the form of a conic frustum surmounted by a cone. The diameters of the base and top of the frustum are 20 m and 6 m respectively and the height is 24 m. If the height of the tent is 28 m, find the quantity of canvas required.

(a) 280π m

(b) 325π m

(c) 235π m

(d) 340π m2

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Answer: (d) = 340π m2  

Area of canvas required = Lateral surface area of the conic frustum + Curved surface area of the cone.

Height of conical part = 28 m – 24 m = 4 m 

Slant height (l1) of the conical part = \(\sqrt{4^2+3^2}\)  = \(\sqrt{25}\) = 5m.

Slant height (l2) of the frustum 

\(\sqrt{h^2+(R-r)^2}\)  = \(\sqrt{24^2+(10-3)^2}\)  

\(\sqrt{576+49} = \sqrt{625}\)  = 25 m.

∴  Area of canvas reqd = π(R + r)l2 + πrl1

= π× 13 × 25 + π × 3 × 5 

= (325π + 15π) m2 

= 340π m2.

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