A cone has a height of #24 cm# and its base has a radius of #6 cm#. If the cone is horizontally cut into two segments #12 cm# from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2018-06-28T11:35:00

Total surface area of bottom segment is #491.11# sq.cm.

The cone is cut at #12# cm from base, So upper radius of the

frustum of cone is #r_2=(24-12)/24*6=3# cm ; slant ht:

#l=sqrt(12^2+(6-3)^2)=sqrt(153)~~ 12.37# cm.

Top surface area #A_t=pi*3^2~~ 28.274# sq.cm

Bottom surface area #A_b=pi*6^2~~113.1 # sq.cm

Slant Area #A_s=pi*l*(r_1+r_2)=pi*12.37*(6+3)#

#~~349.74# sq.cm.

Total surface area of bottom segment

#=A_t+A_b+A_s=28.27+113.1+349.74~~491.11 (2dp)#

sq.cm [Ans]