Question from RS Aggarwal book, problem number 21, page number 824, exercise 17C, chapter volume and surface area of solid.
Chapter Mensuration
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
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