A cone has a height of #24 cm# and its base has a radius of #5 cm#. If the cone is horizontally cut into two segments #15 cm# from the base, what would the surface area of the bottom segment be?

Answer 1

#420.5 cm^2#

Solution: Subtract the top cut outer area from the original to derive the bottom section outer area. Calculate the new circular (cut) section area and the base area. Combine the three values to the final area of the bottom segment.

A horizontal cut means simply that the cone top now has a height of 9cm. The original cone had an angle with a tangent of #24/5#. Therefore, the new lengths must have an equal tangent of #9/r_2#. So, #r_2 = (9/24)*5 = 1.875cm #
The area of the circle is #pi * (1.875)^2 = 11.04# The area of the original base is # pi * (5)^2 = 78.5# The original area of the cone exterior was #pi*r_1*s_1 (sides) + pi*r_1^2#.
From our tangent calculation we know the angle is 78.2’. #sin 78.2 = 24/s = 0.979 ; s_1 = 24.5cm # The original side area was therefore #pi*5*24.5 = 385.1# After the cut, the side length of the top is #9/s = 0.979 ; s_2 = 9.19 # Therefore, the bottom side length is #15.3cm# . The remaining (top) cone after the cut has an exterior side area of #pi*r_2*s_2 # = #pi*1.875*9.19 = 54.1#
Subtracting this from the original exterior side area #385.1 – 54.1 = 331# left on the bottom exterior sides. The circular area at the cut is #pi*r_2^2 = pi*1.875^2 = 11#. Adding this to the previously calculated areas for the sides and bottom circular parts we finally arrive at: 331 (sides) + 78.5 (base) + 11 (top section) = #420.5 cm^2#
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Answer 2

To find the surface area of the bottom segment of the cone, we can use the concept of similar triangles and the formula for the lateral surface area of a cone.

First, we need to find the slant height of the bottom segment of the cone. This can be done using the Pythagorean theorem:

[ \text{Slant height} = \sqrt{(\text{Height of cone})^2 - (\text{Radius of base})^2} ] [ \text{Slant height} = \sqrt{(24 , \text{cm})^2 - (5 , \text{cm})^2} ] [ \text{Slant height} = \sqrt{576 , \text{cm}^2 - 25 , \text{cm}^2} ] [ \text{Slant height} = \sqrt{551 , \text{cm}^2} ] [ \text{Slant height} = 23.46 , \text{cm} ]

Now, we can find the radius of the bottom segment using similar triangles. The ratio of the height of the bottom segment to the height of the whole cone is equal to the ratio of the radius of the bottom segment to the radius of the whole cone. Therefore:

[ \frac{\text{Height of bottom segment}}{\text{Height of cone}} = \frac{\text{Radius of bottom segment}}{\text{Radius of cone}} ] [ \frac{15 , \text{cm}}{24 , \text{cm}} = \frac{\text{Radius of bottom segment}}{5 , \text{cm}} ] [ \text{Radius of bottom segment} = \frac{15}{24} \times 5 ] [ \text{Radius of bottom segment} = 3.125 , \text{cm} ]

Now that we have the slant height and the radius of the bottom segment, we can find its lateral surface area using the formula for the lateral surface area of a cone:

[ \text{Lateral surface area} = \pi \times \text{Radius} \times \text{Slant height} ] [ \text{Lateral surface area} = \pi \times 3.125 , \text{cm} \times 23.46 , \text{cm} ] [ \text{Lateral surface area} \approx 231.51 , \text{cm}^2 ]

Therefore, the surface area of the bottom segment of the cone would be approximately 231.51 square centimeters.

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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.

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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