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

A square base has area $s^2$. The prism’s volume is base area times $2s$; the pyramid’s is one-third base area times its vertical height, $\frac34s$. The pieces add to the given total volume, so

Type $x_1^2(2x_1)+\frac13x_1^2(\frac34x_1)\sim1152\{x_1>0\}$, using $x_1$ for $s$. The $\sim$ tells Desmos to fit the unknown, and the restriction keeps it positive, as a length must be. Under PARAMETERS, Desmos shows $x_1=8$, so $s=8$ centimeters.

Count only faces exposed on the finished solid. The prism has four side rectangles, each $8$ by $16$ because its height is $2s$, plus its bottom square. Its top square is hidden under the pyramid. Type $4(8)(16)+8^2$; Desmos gives $576$ square centimeters.

A slant height $\ell$ runs from the tip to the midpoint of one base edge. Because this is a right square pyramid, the tip is above the square’s center. The vertical leg is $\frac34(8)=6$, and the horizontal leg from center to edge midpoint is $\frac82=4$. These legs meet at a right angle, so the Pythagorean theorem gives the face height, not the vertical height:

Type $6^2+4^2$. Desmos shows $52$, so the right-triangle equation becomes $\ell^2=52$.

A length is positive, so take the positive square root: $\ell=\sqrt{52}$. Rewrite $52$ as $4\cdot13$: $\ell=\sqrt{4\cdot13}$. Pull the perfect square $4$ out of the radical:

Each exposed triangle has area $\frac12s\ell$. For four faces, type the coefficient $4(\frac12)(8)(2)$; Desmos gives $32$. Keep the radical from $\ell=2\sqrt{13}$, so the pyramid contributes $32\sqrt{13}$ square centimeters.

Type $576+32\sqrt{13}$ to add the prism’s area to the pyramid’s area; Desmos shows about $691.38$. The total exterior surface area is exactly $576+32\sqrt{13}$ square centimeters. Choice B.