Question

A cone has a height of `16 cm` and its base has a radius of `3 cm`. If the cone is horizontally cut into two segments `10 cm` from the base, what would the surface area of the bottom segment be?

Answer 1

Total surface area of bottom segment is `~~164.10 (2dp)` sq.cm

Explanation:

The cone is cut at `10` cm from base, So upper radius of the

frustum of cone is `r_2=(16-10)/16*3=1.125` cm ; Slant ht:

`l=sqrt(10^2+(3-1.125)^2)~~10.174`

Top surface area `A_t=pi*1.125^2 ~~3.976` sq.cm

Bottom surface area `A_b=pi*3^2~~28.274` sq.cm

Slant Area: `A_s=pi*l*(r_1+r_2)=pi*10.174*(3+1.125)`

`~~131.849` sq.cm. Total surface area of bottom segment is

`A_t+A_b+A_s=3.976+28.274+131.849`

`~~164.10 (2dp)`sq.cm [Ans]