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Answer :
To wrap 8 container 794.42 in² of plastic wrap required.
we have,
Diameter = 5.5 inch
radius= 5.5/2 inch
and, height = 3 inch
Surface area of Cylinder
= 2πrh + 2πr²
= 2 x 3.14 x 5.5/2 x 3 + 2 x 3.14 x 5.5/2 x 5.5/2
= 51.81 + 47.4925
= 99.3325 in²
Now, to wrap 8 container the amount of plastic wrap required
= 8 x 99.3325
= 794.42 in²
Learn more about Surface area here:
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