Question

A cone has a height of `9 cm` and its base has a radius of `2 cm`. If the cone is horizontally cut into two segments `4 cm` from the base, what would the surface area of the bottom segment be?

Answer 1

To see an in-depth explanation of the answer, see
https://socratic.org/questions/a-cone-has-a-height-of-16-cm-and-its-base-has-a-radius-of-3-cm-if-the-cone-is-ho-1#478093 and https://socratic.org/questions/a-cone-has-a-height-of-16-cm-and-its-base-has-a-radius-of-8-cm-if-the-cone-is-ho-1#476669

`R_2` equals
`9/2=4/R_2`
`6=9R_2`
`R_2=6/9 =2/3`

`s` equals
`s=sqrt((2-2/3)^2+16)`
`s=sqrt((1 1/3)^2+16)`
`s=sqrt(160/9)`

Surface area equals
`A_s=pi(sqrt(160/9)(2+2/3)+2^2+(2/3)^2)`
`A_s=pi((16sqrt10)/9+4+(4/9))`
`A_s=pi((8(5+2sqrt40))/9)`
`A_s~~31.624cm^2`

Therefore the surface area of the bottom segment of the cone is roughly 31.624`cm^2`.

I hope I helped!