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 $\pi = 22/7$, the volume, in cubic ft, of the remaining part of the cone is

We have a right circular cone (imagine an ice cream cone standing upright):

Height: 12 ft

Base diameter: 8 ft, so radius = 4 ft

A horizontal cut is made 9 ft from the base (which means 3 ft from the top)

We need to find the volume of the remaining part after the tip is removed.

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Instead of trying to calculate the volume of the weird remaining shape directly, we'll use this clever method:

Volume of remaining part = Volume of original cone - Volume of cut-off tip

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When we cut a cone with a plane parallel to the base, the cut-off piece is also a cone similar to the original.

Using similar triangles:

Original cone: height 12 ft, radius 4 ft

Cut-off tip: height 3 ft, radius = ?

Since the triangles are similar:

$\tfrac{\text{radius of tip}}{\text{height of tip}} = \tfrac{\text{radius of original}}{\text{height of original}}$

$\tfrac{r_{tip}}{3} = \tfrac{4}{12} = \tfrac{1}{3}$

Therefore: $r_{tip} = 3 \times \tfrac{1}{3} = 1 \text{ ft}$

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The volume of any cone is: $V = \tfrac{1}{3}\pi r^2 h$

Volume of original cone:

$V_{original} = \tfrac{1}{3}\pi (4)^2 (12) = \tfrac{1}{3}\pi (16)(12) = \tfrac{192\pi}{3} = 64\pi$

Volume of cut-off tip:

$V_{tip} = \tfrac{1}{3}\pi (1)^2 (3) = \tfrac{1}{3}\pi (1)(3) = \tfrac{3\pi}{3} = \pi$

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$V_{remaining} = V_{original} - V_{tip} = 64\pi - \pi = 63\pi$

Substitute $\pi = \tfrac{22}{7}$:

$V_{remaining} = 63 \times \tfrac{22}{7} = \tfrac{63 \times 22}{7} = \tfrac{1386}{7} = 198 \text{ cubic ft}$

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When dealing with truncated cones (cones with tops cut off), we always use similarity to find the dimensions of the cut-off piece, then subtract volumes. This method is much faster than trying to use complex formulas for truncated cone volumes directly.

Final Answer: 198 cubic ft