Question

A cone has a height of `36 cm` and its base has a radius of `9 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?

Answer 1

Total surface area of bottom segment is `950.46` sq.cm

Explanation:

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

of cone is `r_2=(36-12)/36*9=6` cm ; slant height of the frustum

is `l=sqrt(12^2+(9-6)^2)=sqrt(144+9)=sqrt 153=12.37` cm

Top surface area `A_t=pi*6^2=113.1` sq.cm

Bottom surface area `A_b=pi*9^2=254.47`sq.cm

Slant area `A_s=pi*l*(r_1+r_2)=pi*12.37*(9+6)=582.89`sq.cm

Total surface area of bottom segment

`=A_t+A_b+A_s=113.1+254.47+582.89=950.46 (2dp)` sq.cm [Ans]