An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, then find the area of the tin sheet required to make the funnel

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Question from RS Aggarwal book, problem number 21, page number 824, exercise 17C, chapter volume and surface area of solid.

solution

given data

Height of the cylindrical portion, h=10 cm

Height of the frustum of cone portion, H=22-10=12 cm

Radius of the cylindical portion=Radius of smaller end of frustum portion, r=8/2=4

Radius of larger end of frustum portion, R=182=9 cm

the slant height of the frustum,

l= √[[R-r]²+H²]

= √[[9-4]²+12²]

= √[5²+12]²

=√[25+144]

=√169

l =13 cm

The area of the tin sheet required= CSA of frustum part + CSA of cylindrical part

=π[R+r]l+2πrh

=[22/7]×[9+4]×13 + [2×[22/7]×4×10

=[22/7]×13×13 + [[22/7]×80

=[22/7]×169+80

=[22/7]×249

= 782.57 cm²

∴ the area of the tin sheet required to make the funnel is 782.57 cm²