Question

A cone has a height of `9 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 `458.55(2dp)` sq.cm

Explanation:

The cone is cut at 3 cm from base, So upper radius of the frustum of cone is `r_2=(9-3)/9*8=16/3`cm ;

Slant ht `l=sqrt(3^2+(8-16/3)^2)=sqrt(9+64/9)=sqrt (145/9)=4.01`cm

Top surface area `A_t=pi*(16/3)^2=89.36` 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*4.01*(8+16/3)= 168.13` sq.cm

Total surface area of bottom segment `=A_t+A_b+A_s=89.36+201.06+168.13=458.55(2dp)`sq.cm [Ans]