# 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?

Mar 8, 2018

$S . A . = 196 \pi$ $c {m}^{2}$

#### 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 $r = 8$ $c m$ explicitly, whereas we would let $h$ be $4$ $c m$ since the question is asking for $S . A .$ of the bottom cylinder.
$S . A . = 2 \pi \cdot {r}^{2} + 2 \pi \cdot r \cdot h = 2 \pi \cdot r \cdot \left(r + h\right)$

Plug in the numbers and we get:
$2 \pi \cdot \left({8}^{2} + 8 \cdot 4\right) = 196 \pi$

Which is approximately $615.8$ $c {m}^{2}$.

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 $2 \pi \cdot r$. (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 $\pi \cdot d = 2 \pi \cdot r$.)

Now we find the area formula for each of the component: ${A}_{\text{circle}} = \pi \cdot {r}^{2}$ for each of the circle, and ${A}_{\text{rectangle}} = h \cdot l = h \cdot \left(2 \pi \cdot r\right) = 2 \pi \cdot r \cdot h$ for the rectangle.

Adding them to find an expression for the surface area of the cylinder:
$S . A . = 2 \cdot {A}_{\text{circle"+A_"rectangle}} = 2 \pi \cdot {r}^{2} + 2 \pi \cdot r \cdot h$
Factor out $2 \pi \cdot r$ to get $S . A . = 2 \pi \cdot r \cdot \left(r + h\right)$

Notice that since each cylinder has two caps, there are two ${A}_{\text{circle}}$ *in the expression for * $S . A .$

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.

Apr 15, 2018

:.color(purple)(=491.796cm^2 to the nearest 3 decimal places $c {m}^{2}$

#### Explanation:

:.Pythagoras: ${c}^{2} = {12}^{2} + {8}^{2}$

$\therefore c = L = \sqrt{{12}^{2} + {8}^{2}}$

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

:.12/8=tan theta=1.5=56^@18’35.7”

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

:.S.A.$= \pi \cdot 8 \cdot 14.422$

:.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: ${c}^{2} = {8}^{2} + {5.333}^{2}$

$\therefore c = L = \sqrt{{8}^{2} + {5.333}^{2}}$

:. c=Lcolor(purple)(=9.615cm top part
:.S.A. top part$= \pi \cdot r \cdot L$

S.A. top part$\therefore \pi \cdot 5.333 \cdot 9.615$

S.A. top part$\therefore = 161.091$

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

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

:.S.A. Bottom part$= 201.373 + 89.361 + 201.062 = 491.796 c {m}^{2}$
:.color(purple)(=491.796cm^2# to the nearest 3 decimal places $c {m}^{2}$