An oil funnel made of tin sheet consists of a 10 cm long cylindr

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444.

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, find the area of the tin sheet required to make the funnel (see the given Fig.


Let R and r be respectively the radii of the bigger and smaller ends of the frustum, then

straight R space equals space 18 over 2 equals 9 space cm semicolon space space space straight r space equals space 8 over 2 equals 4 space cm
Let l and h be respectively the slant height and height of frustum, then
h = total height – height of cylindrical part
= 22 cm – 10 cm
= 12 cm
 a n d space space space space space space space space l italic space italic space italic space equals space space italic space square root of straight h squared plus left parenthesis straight R minus straight r right parenthesis squared end root italic space
italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space equals space space space square root of left parenthesis 12 right parenthesis squared plus left parenthesis 9 minus 4 right parenthesis squared end root
space space space space space space space space space space space space space space space space equals space space space space square root of 144 plus 25 end root
space space space space space space space space space space space space space space space space equals space space space space square root of 169 space equals space 13 space cm

Now,
Curved surface area of frustum

equals space rl space left parenthesis straight R space plus space straight r right parenthesis
equals space 22 over 7 straight x space 13 space left parenthesis 9 plus 4 right parenthesis
equals space space 22 over 7 straight x space 13 space straight x space 13
equals space 531.14 space cm squared

Let r1 and h1 be respectively the radius and height of cylindrical part.
Then, r1 = 4cm and h1 = 10cm
Now,
Curved Surface area of the cylinder
= 2πr11

equals space open parentheses 2 space straight x space 22 over 7 straight x space 4 space straight x space 10 close parentheses space cm squared
equals space 251.43 space cm squared

Hence,
Area of tin required
= Curved surface area of frustum + curved surface area of cylinder
= 531.14 + 251.43
= 782.57 cm2.

 

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