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An oil funnel made of tin sheet consists of a 10 cm long cylinder 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 the tin sheet required to make funnel.
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Let R and r be respectively the radii of bigger and smaller ends of frustum
then,
R=18/2=9, r=8/2=4
Let l and h be respectively the slant height and height of frustum
then,
h=total height-height of cylindrical part
=22-1 0
=12
and l= √h×h+(R-r)(R-r)
=√(12×12)+(9-4)(9-4)
=√144+25
=√169=13
Now
Curved surface area of frustum
rl(R+r)
22/7×13×(9+4)
22/7×13×13
531.14cm sq.
Let r1 and h respectively the radius and height of cylindrical part
r1=4 cm.
h1=10 cm.
now
curved surface area
2×22/7×4×10
251.43 cm sq
Area of tin
531.14+251.43
782.57 cm sq.
Mark it as brainliest if it helps you please!!!
then,
R=18/2=9, r=8/2=4
Let l and h be respectively the slant height and height of frustum
then,
h=total height-height of cylindrical part
=22-1 0
=12
and l= √h×h+(R-r)(R-r)
=√(12×12)+(9-4)(9-4)
=√144+25
=√169=13
Now
Curved surface area of frustum
rl(R+r)
22/7×13×(9+4)
22/7×13×13
531.14cm sq.
Let r1 and h respectively the radius and height of cylindrical part
r1=4 cm.
h1=10 cm.
now
curved surface area
2×22/7×4×10
251.43 cm sq
Area of tin
531.14+251.43
782.57 cm sq.
Mark it as brainliest if it helps you please!!!
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