Question 19·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

The apex is at the cube’s center, halfway from the face with the opening to the opposite face. So the perpendicular height, measured straight from the apex to the opening, is $h=\frac{18}{2}=9\text{ centimeters}$.

Let $\ell$ be the slant height from the apex to the midpoint of one opening edge. The perpendicular height $9$ and the distance $x/2$ from the opening’s center to that midpoint form a right triangle. The Pythagorean theorem gives: $\ell=\sqrt{9^2+\left(\frac{x}{2}\right)^2}$. Use this slant height for wall area, not the perpendicular height $9$.

Each of the four triangular walls has area $\frac12x\ell$. The square opening of area $x^2$ is no longer on the surface, so the given increase is new wall area minus removed square area: $4\left(\frac12x\ell\right)-x^2=108$. Combine the four equal wall areas: $2x\ell-x^2=108$. Counting only the walls would make the increase too large.

Substitute the slant height into the area equation. In Desmos, type $2x_1\sqrt{9^2+(x_1/2)^2}-x_1^2\sim108\{0<x_1\le18\}$. A regression uses $x_1$ and $\sim$ to find a value that makes the two sides equal; the restriction keeps the side positive and no longer than the cube’s edge. Under PARAMETERS, Desmos shows $x_1\approx10.3923$ centimeters.

The pyramid’s square base has area $x^2$. Add $x_1^2$ on the next Desmos line; it shows $108$ square centimeters. Desmos uses the full fitted value of $x_1$, rather than the shortened decimal shown under PARAMETERS.

A pyramid’s volume is one-third its base area times its perpendicular height. Subtract the carved-out volume from the original volume. So the remaining volume is $V=18^3-\frac13x^2(9)$. Add $18^3-\frac13x_1^2(9)$ in Desmos. It shows $5508$, the resulting solid’s volume in cubic centimeters. Grid in 5508.