Question 12·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

Let $p$ mean the small pyramid’s volume divided by the whole pyramid’s volume. If the small pyramid has $8$ equal volume parts, the frustum has $19$, so the whole has $8+19$. The whole, not the frustum, belongs in the denominator: $p=\frac{8}{8+19}$. Type $p=\frac{8}{8+19}$ in Desmos. It shows $p\approx0.296296$; the fraction button shows $p=\frac8{27}$.

The plane is parallel to the base, so the top and whole pyramids are similar: matching lengths have the same factor $k$. Volume scales by $k^3$, but area scales by $k^2$. So $k^3=p$. Take the positive cube root, since lengths are positive: $k=p^{1/3}$. Type $k=p^{1/3}$. Desmos shows $0.666\ldots$; its fraction button shows $\frac23$. That’s the top-to-whole length factor.

For a pyramid, $V=\frac13Bh$, where $B$ is its base area and $h$ is its perpendicular height. The given height and volume belong to the whole pyramid. Type $\frac13x_1(24)\sim10368$, with $x_1$ standing for $B$; $\sim$ tells Desmos to find the value that fits. Under PARAMETERS, it shows $x_1=1296$, so the original base area is $1{,}296$ square inches.

The cross section is the smaller pyramid’s square base. Its area uses $k^2$, not $k$, because both side lengths shrink: $A_{\text{cut}}=k^2B$. Type $k^2x_1$ as the last Desmos line; it prints $576$. The cross section has an area of $576$ square inches. Grid in 576.