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A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With π = 22/7, the volume, in cubic ft, of the remaining part of the cone is

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To calculate the volume of the remaining part of the cone, we need to find the volume of the original cone and subtract the volume of the smaller cone that was cut off.

Let's start by finding the volume of the original cone. The formula for the volume of a cone is V = (1/3) * π * r^2 * h, where r is the radius of the base and h is the height.

Given that the diameter of the base is 8 ft, the radius (r) is half of that, which is 4 ft. The height (h) of the original cone is 12 ft. We'll use the value of π as 22/7.

V_original = (1/3) * (22/7) * (4 ft)^2 * 12 ft
V_original = (1/3) * (22/7) * 16 ft^2 * 12 ft
V_original = (22/21) * 192 ft^3
V_original ≈ 704 ft^3

Now, let's find the volume of the smaller cone that was cut off. The height of the smaller cone is 9 ft, and the radius is the same as the original cone, which is 4 ft.

V_cut_off = (1/3) * (22/7) * (4 ft)^2 * 9 ft
V_cut_off = (1/3) * (22/7) * 16 ft^2 * 9 ft
V_cut_off = (22/21) * 144 ft^3
V_cut_off ≈ 176 ft^3

Finally, to find the volume of the remaining part of the cone, we subtract the volume of the cut-off portion from the volume of the original cone:

V_remaining = V_original - V_cut_off
V_remaining = 704 ft^3 - 176 ft^3
V_remaining = 528 ft^3

Therefore, the volume of the remaining part of the cone is approximately 528 cubic ft.
User Onedayitwillmake
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