Question 24·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

The base is a circle, so its area is $\pi r^2$. The given area gives:

Type $225\pi\sim\pi r^2\{r>0\}$ in Desmos. The $\sim$ asks Desmos to fit $r$, and the restriction keeps this radius positive. Under PARAMETERS, Desmos shows $r=15$ centimeters.

The cut is parallel to the base, so the two cones are similar: every matching length has the same factor $k$ from the original cone to the smaller one. Volume uses three lengths, so volume scales by the cube of the length factor:

Type $\frac{8}{125}\sim k^3\{k>0\}$ in Desmos. Under PARAMETERS, it shows $k=0.4$, which is $\frac25$. Using $\frac{8}{125}$ directly on a radius would make the cone too small.

The radius $15$ and vertical height $20$ meet at a right angle. The slant height $\ell$ runs along the cone’s side, so it is the third side of that triangle. Use the $15$-$20$-$25$ right-triangle relationship:

So $\ell=25$ centimeters. The vertical height is not the length used for curved surface area.

Apply the same factor $\frac25$ to each matching length. Scale the radius:

Scale the slant height:

These are the smaller cone’s radius and slant height, which locate the new top circle and the curved side removed.

A cone’s curved area is $\pi r\ell$. Subtract the smaller cone’s curved area from the original’s, then add the original bottom circle and the new exposed top circle. This counts the finished frustum’s surfaces:

Subtracting the smaller cone’s entire surface would wrongly subtract its circular base instead of adding the new top.

Type $15\cdot25-6\cdot10+15^2+6^2$ in Desmos. It prints $576$, the number multiplying $\pi$. The remaining frustum has a total surface area of $576\pi$ square centimeters. Choice D.