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?

2 Answers
Mar 8, 2018

#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.
Unrolled cylinder diagram- CK-12 Foundation

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.

Apr 15, 2018

#:.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#