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 $5 cm$ from the base, what would the surface area of the bottom segment be?

Question

1749 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-19T12:21:45

Total surface area of bottom segment is 358.04 (2dp)# sq.cm.

The cone is cut at 5 cm from base, So upper radius of the frustum of

cone is $r_2=(24-5)/24*6=4.75$ cm ; slant ht:

$l=sqrt(5^2+(6-4.75)^2)=sqrt(25+1.5625)=sqrt 26.5625=5.15$

Top surface area $A_t=pi*4.75^2=70.88$ sq.cm

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

Slant Area $A_s=pi*l*(r_1+r_2)=pi*5.15*(6+4.75)=174.06$ sq.cm

Total surface area of bottom segment

$=A_t+A_b+A_s=70.88+113.1+174.06=358.04 (2dp)$

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 5 cm5 cm from the base, what would the surface area of the bottom segment be?