Question

A cone has a height of `12 cm` and its base has a radius of `8 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

`S.A.=196pi` `cm^2`

Explanation:

Apply the formula for the surface area (`S.A.`) of a cylinder with height `h` and base radius `r`. The question has stated that `r=8` `cm` explicitly, whereas we would let `h` be `4` `cm` since the question is asking for `S.A.` of the bottom cylinder.
`S.A.=2pi*r^2+2pi*r*h=2pi*r*(r+h)`

Plug in the numbers and we get:
`2pi*(8^2+8*4)=196pi`

Which is approximately `615.8` `cm^2`.

You might think about this formula by imaging the products of an exploded (or unrolled) cylinder.

The cylinder would include three surfaces: a pair of identical circles of radii of `r` that act as caps, and a rectangular wall of height `h` and length `2pi*r`. (Why? Since when forming the cylinder the very rectangle would roll into a tube, precisely matching the outer rim of both circles that have circumferences `pi*d=2pi*r`.)

Now we find the area formula for each of the component: `A_"circle"=pi*r^2` for each of the circle, and `A_"rectangle"=h*l=h*(2pi*r)=2pi*r*h` for the rectangle.

Adding them to find an expression for the surface area of the cylinder:
`S.A.=2*A_"circle"+A_"rectangle"=2pi*r^2+2pi*r*h`
Factor out `2pi*r` to get `S.A.=2pi*r*(r+h)`

Notice that since each cylinder has two caps, there are two `A_"circle"` *in the expression for * `S.A.`

Reference and Image Attributions:
Niemann, Bonnie, and Jen Kershaw. “Surface Area of Cylinders.” CK-12 Foundation, CK-12 Foundation, 8 Sept. 2016, www.ck12.org/geometry/surface-area-of-cylinders/lesson/Surface-Area-of-Cylinders-MSM7/?referrer=concept_details.

Answer 2

`:.color(purple)(=491.796cm^2` to the nearest 3 decimal places `cm^2`

Explanation:

:.Pythagoras: `c^2=12^2+8^2`

`:.c=L=sqrt(12^2+8^2)`

`:. c=Lcolor(purple)(=14.422cm`

`:.12/8=tan theta=1.5=56^@18’35.7”`

:.`color(purple)(S.A.`= pirL#

:.S.A.`=pi*8*14.422`

:.S.A.`=362.464`

:.Total S.A.`color(purple)(=362.464cm^2`

`:.Cot 56^@18’35.7”*8=5.333cm=`radius of top part

:.Pythagoras: `c^2=8^2+5.333^2`

`:.c=L=sqrt(8^2+5.333^2)`

`:. c=Lcolor(purple)(=9.615cm` top part
:.S.A. top part`=pi*r*L`

S.A. top part`:.pi*5.333*9.615`

S.A. top part`:.=161.091`

S.A. top part`:.color(purple)(=161.091cm^2`

:.S.A. Bottom part`color(purple)(=362.464-161.091=201.373cm^2`

:.S.A. Bottom part`=201.373+89.361+201.062=491.796 cm^2`
`:.color(purple)(=491.796cm^2` to the nearest 3 decimal places `cm^2`