Question

A cone has a height of `16 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 `486.58(2dp)` sq,cm

Explanation:

The cone is cut at 3 cm from base, So upper radius of the frustum of cone is `r_2=(16-3)/16*8=6.5`cm ; slant ht `l=sqrt(3^2+(8-6.5)^2)=sqrt(9+2.25)=sqrt 11.25=3.354 cm`

Top surface area `A_t=pi*6.5^2=132.73` cm^2
Bottom surface area `A_b=pi*8^2=201.06`cm^2
Slant Area `A_s=pi*l*(r_1+r_2)=pi*3.354*(8+6.5)=152.79`cm^2
Total surface area of bottom segment `=A_t+A_b+A_s=132.73+201.06+153.79=486.58(2dp)`sq,cm[Ans]