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 `16 cm` from the base, what would the surface area of the bottom segment be?

Answer 1

Total surface area of bottom segment is `1058.38(2dp)` sq.cm

Explanation:

The cone is cut at 16 cm from base, So upper radius of the frustum of cone is `r_2=(36-16)/36*9=5` cm ; slant ht `l=sqrt(16^2+(9-5)^2)=sqrt(256+16)=sqrt 272=16.49` cm

Top surface area `A_t=pi*5^2=78.54` 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*16.49*(9+5)=725.37` sq.cm

Total surface area of bottom segment `=A_t+A_b+A_s=78.54+254.47+725.37=1058.38(2dp)`sq.cm[Ans]