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#

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#

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Answer 2

To find the surface area of the bottom segment of the cone, first calculate the slant height of the bottom segment using the Pythagorean theorem. Then, use the formula for the lateral surface area of a cone to find the surface area of the bottom segment.

Given: Height of cone (h) = 5 cm Radius of cone base (r) = 5 cm Distance from base to cut (d) = 3 cm

Using the Pythagorean theorem: [l = \sqrt{r^2 + (h - d)^2}] [l = \sqrt{5^2 + (5 - 3)^2}] [l = \sqrt{25 + 4}] [l = \sqrt{29}]

The lateral surface area of the cone (excluding the base) is: [A_{\text{lateral}} = \pi r l] [A_{\text{lateral}} = \pi \times 5 \times \sqrt{29}]

The surface area of the bottom segment is then: [A_{\text{bottom segment}} = A_{\text{lateral}} + \pi r^2] [A_{\text{bottom segment}} = \pi \times 5 \times \sqrt{29} + \pi \times 5^2]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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