Question

A cone has a height of `24 cm` and its base has a radius of `8 cm`. If the cone is horizontally cut into two segments `3 cm` from the base, what would the surface area of the bottom segment be?

Answer 1

Total surface area of bottom segment is `504.02 (2dp)` sq.cm

Explanation:

The cone is cut at 3 cm from base, So upper radius of the frustum

of cone is `r_2=(24-3)/24*8=7` cm ;

Slant ht: `l=sqrt(3^2+(8-7)^2)=sqrt(9+1)=sqrt 10~~3.16` cm.

Top surface area `A_t=pi*7^2 ~~153.94` sq.cm

Bottom surface area `A_b=pi*8^2~~201.06` sq.cm

Slant Area `A_s=pi*l*(r_1+r_2)=pi*3.16*(8+7)~~149.02` sq.cm.

Total surface area of bottom segment is

`=A_t+A_b+A_s=153.94+201.06.1+149.02~~504.02 (2dp)`

sq.cm [Ans]