Question

A cone has a height of `12 cm` and its base has a radius of `6 cm`. If the cone is horizontally cut into two segments `6 cm` from the base, what would the surface area of the bottom segment be?

Answer 1

Total surface area of bottom segment is `331.04(2dp) cm^2`

Explanation:

The cone is cut at 6 cm from base, So upper radius of the frustum of cone is `r_2=(12-6)/12*6=3 cm` ; slant ht `l=sqrt(6^2+(6-3)^2)=sqrt(36+9)=sqrt 45=6.71(2dp) cm`

Top surface area `A_t=pi*3^2=28.27 cm^2`
Bottom surface area `A_b=pi*6^2=113.10 cm^2`
Slant Area `A_s=pi*l*(r_1+r_2)=pi*6.71*(6+3)=189.67 cm^2`

Total surface area of bottom segment `=A_t+A_b+A_s=28.27+113.10+189.67=331.04(2dp) cm^2`[Ans]