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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 $\pi = 22/7$, the volume, in cubic ft, of the remaining part of the cone is ________
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We can draw the circular cone.



Here, $ \triangle \text{ADE} \sim \triangle \text{ABC} $

So, $ \frac{ \text{AD}}{ \text{AB}} = \frac{ \text{DE}}{ \text{BC}} $

$ \Rightarrow \frac{3}{12} = \frac{ \text{DE}}{4}$

$ \Rightarrow \boxed{\text{DE} = 1 \; \text{ft}}$

The volume of the cone $ = \frac{1}{3} \pi r^{2}h\; ;$ where $`r\text{’}$ is the base radius of the cone, and $`h\text{’}$ is the height of the cone.

Now, the volume of big cone $ = \frac{1}{3} \pi \times 4^{2} \times 12 = 64 \pi \; \text{cubic ft} $

The volume of small cone $ = \frac{1}{3} \pi \times 1^{2} \times 3 = \pi \; \text{cubic ft} $

$\therefore$ The required volume of the remaining part of the cone $ = $ the volume of big cone $–$ the volume of small cone $ = 64 \pi – \pi = 63 \pi = 63 \times \frac {22}{7} = 198 \; \text{ cubic ft}.$

Correct Answer $: 198$

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