Question 72·Hard·Math

Let $k$ be the ratio of the small cone’s height to the big cone’s height. For similar solids, how does the volume scale in terms of $k$?

The volume of the portion that remains is the original cone’s volume minus the small cone’s volume. Write this in terms of $k^3$ and set it equal to $\tfrac{7}{8}$ of the original volume.

Once you know $k$, use $k=\dfrac{h}{15}$ to find $h$, the height of the small cone.

Enter $y=1-k^3$ and $y=\frac{7}{8}$ on separate expression lines in Desmos. Select their intersection and read the $k$-coordinate, which is the linear scale factor.

Enter $15k$ using the value of $k$ from the intersection. The resulting value is the height of the removed cone in centimeters.

Because the plane is parallel to the base, the removed top cone is similar to the original cone.

Let $H=15$ be the height of the original cone, and let $h$ be the height of the removed cone. Define the linear scale factor as

Since the removed cone is smaller, $0<k<1$.

For similar three-dimensional figures, volumes scale by the cube of the linear scale factor. If $V$ is the volume of the original cone and $V_s$ is the volume of the removed cone, then

The portion that remains has volume

The remaining portion has volume $\tfrac{7}{8}V$, so

Therefore,

Since $k=\frac{h}{15}$,

Thus,

The removed cone's height is $7.5$ centimeters.