Question

A cone has a height of `5 cm` and its base has a radius of `5 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

`SA_("frustum") = (14+21sqrt(2))pi`
`" " ~~ 137.28 \ cm^2`

Explanation:

Consider a cross section:

We want to calculate the surface area of the base (after a cone section is remove), also known as a frustum .

Here we have `AF=5, DF=5, FG=3 => AG=2\ \` (cm)

Firstly we can calculate the curved surface area of the original cone `AED` using `SA=pirl` is:

`SA_(AED) = pi * DF * AD`

We can calculate `AD` using Pythagoras:

`AD^2 = AF^2+DF^2`
`" " = 5^2+5^2`
`" " = 25+25`
`" " = 50 => AD = sqrt(50)=5sqrt(2)`

And so returning to the curved Surface Area, we get:

`SA_(AED) = pi * 5 * 5sqrt(2)`
`" " = 25sqrt(2)pi`

We can perform a similar calculation for the upper cone `ABC`

`AB^2=BG^2+AG^2`

We are not given the upper radius if the frustum, we can however use similar triangles to calculate it:

`triangle AEF` is similar to `triangle ACG`
`:. (AF)/(EF) = (AG)/(CG)`
`:. (5)/(5) = (2)/(CG) => CG=BG=2`

Therefore returning to our Pythagorean equation we have:

`AB^2=2^2+2^2`
`" "=4+4`
`" "=8 => AB=sqrt(8)=2sqrt(2)`

And so we can calculate the curved surface area of the upper cone `ABC` using `SA=pirl` is:

`SA_(ABC) = pi * BG * AB`
`" " = pi * 2 * 2sqrt(2)`
`" " = 4sqrt(2)pi`

And so the curved surface area of the frustum is the difference between these calculation:

`SA_(BDCE) = SA_(AED) - SA_(ABC)`
`" " = 25sqrt(2)pi - 4sqrt(2)pi`
`" " = 21sqrt(2)pi`

And the total surface are would also include the circular base and top of frustum (using `2pir`):

`SA_(base) = 2pi*DF`
`" " = 2pi*5`
`" " = 10pi`

`SA_(top) \ \ \ = 2pi*BG`
`" " = 2pi*2`
`" " = 4pi`

Hence the complete surface are of the remaining frustum is:

`SA_("frustum") = SA_(BDCE) + SA_(base) + SA_(top)`
`" " = 21sqrt(2)pi + 10pi+4pi`
`" " = 21sqrt(2)pi + 14pi`
`" " = (14+21sqrt(2))pi`
`" " ~~ 137.28 \ cm^2`