Answer:
Required amount of aluminium foil = 522.80 [tex]in^{2}[/tex]
Step-by-step explanation:
The shape of the large Hershey kisses can be compared to that of a cone. So that;
the surface area of the kisses = [tex]\pi[/tex][tex]r^{2}[/tex] + [tex]\pi[/tex]rl
Let the slant height be represented by l. Applying the Pythagoras theorem;
[tex]l^{2}[/tex] = [tex]10^{2}[/tex] + [tex]8^{2}[/tex]
= 100 + 64
= 164
l = [tex]\sqrt{164}[/tex]
l = 12.81 in
Thus,
surface area of the kisses = (3.14 x [tex]8^{2}[/tex]) + (3.14 x 8 x 12.81)
= 200.96 + 321.7872
= 522.7472
surface area of the kisses = 522.75 [tex]in^{2}[/tex]
Thus, the amount of aluminium foil required to wrap the kisses is 522.80 [tex]in^{2}[/tex].
