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]

A cone has a height of 24 cm24 cm and its base has a radius of 6 cm6 cm. If the cone is horizontally cut into two segments 12 cm12 cm from the base, what would the surface area of the bottom segment be?