Question

A cone is inscribed in a right square pyramid. What is the remaining volume if the cone is removed? 18 16

Answer 1

well, since the cone is touching the edges of the pyramid that means the cone has a diameter of 16 units and thus it has a radius of half that or 8 units, whilst both have a height of 18 units.

Let's simply get the volume of the pyramid and that of the cone and subtract the volume of the cone, what's leftover is the part we didn't subtract, or the part that remains once the cone has been removed.

`\textit{volume of a pyramid}\\\\ V=\cfrac{Bh}{3} ~~ \begin{cases} B=\stackrel{base's}{area}\\ h=height\\[-0.5em] \hrulefill\\ B=\stackrel{16* 16}{256}\\ h=18 \end{cases}\implies V=\cfrac{(256)(18)}{3} \\\\[-0.35em] ~\dotfill`

`\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=8\\ h=18 \end{cases}\implies V=\cfrac{\pi (8)^2(18)}{3} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(256)(18)}{3}~~ - ~~\cfrac{\pi (8)^2(18)}{3}\implies 1536-384\pi ~~ \approx ~~ \text{\LARGE 329.6}`

Answer 2

• Answer:

• Hence, The volume of the remaining:

1536 - 1206.4 = 329.6 Units^3

• Explanation:

The remaining volume is the pyramid Minus (-) the magnitude (volume) of the cone.

• Now, the volume of the pyramid is:

1/3 * 16^2 * 18 = 1536 Units^3

• The volume of a square pyramid is:

1/3 * side^2 * height

• The volume of the cone is:

1/3 * π * (16/2)^2 * 18 = 1206.4 Units ^3

• The volume of a cone is:

1/3 * π * radius^2 * height

• Hence, the volume that is remaining is:

1536 - 1206.4 = 329.6 Units^3

• Answer:

( 329.6 Units^3 )

I hope it helps!