Question

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`

Explanation:

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`