A cone has a height of #6 cm# and its base has a radius of #6 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?

2 Answers
Apr 27, 2016

Yosief you answer is great but you just subtracted the original cone and the top cone. While that will give you a complete solution for the volume for the surface area you have three parts you need to consider:
1) The Base Area => this area of the circle
2) The side Area, that is the one you got by #pis(R-r)# that you have right
3) Because you cut it now you have a smaller circle on top so you need to this area.

So using your terms
#SA_F = "BA + Side Area + Top Area"#
#SA_F = piR^2 + pis(R-r) + pir^2 = pi(R^2+r^2) + pis(R-r)#

What changed? Well in your formula the first is the sum of squares instead the difference of squares. You were not that far you were a sign away from the correct answer. Make that change and voila you got it...

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May 2, 2016

Answer:

Surface area of bottom shape = #9pi(4 + sqrt18)#

Explanation:

Let's start off by writing down what is already known and then finding some of the values we will need later.

The BIG cone and the small cone are similar figures.
R = 6cm ; r = 3cm
H = 6cm ; h = 3cm
L = #sqrt72# ; # l# = #sqrt18# (#l# is the slant height, found by Pythagoras)

Small cone:
surface area = area base + curved surface
s.a. = #pi r^2# + #pi r l# = #pi(3)^2 + pi (3)sqrt18#

Factorise to make it easier: s.a. = #3pi(3 + sqrt18)#

In similar figures, the areas are in the same ratio as the square of the sides, length, radius etc.

#3^2/6^2# = #(3pi(3 + sqrt18))/(SA)# #rArr# #1/4# = #(3pi(3 + sqrt18))/(SA)#

BIG CONE: Surface area = #4 xx 3pi(3 + sqrt18)#
= #12pi(3 + sqrt18)#

Surface area of fustrum (bottom segment);

BIG AREA - small area

= #12pi(3 + sqrt18)# - #3pi(3 + sqrt18)#

= #9pi(3 + sqrt18)#

However, the circular cut surface must be included as well.

Total Surface area = #9pi(3 + sqrt18)# + #pi(3)^2#
= #9pi(3 + sqrt18)# + #9pi#
= #9pi(3 + sqrt18 +1)#
= #9pi(4 + sqrt18)#