Question 35·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

Let $s$ be a base side. A square’s area is its side squared, so $s^2=196$. Type $s_1^2\sim196\{s_1>0\}$ in Desmos. The $\sim$ asks Desmos to find a value, and the restriction keeps the length positive. Under PARAMETERS, it shows $s_1=14$, so each base side is $14$ inches.

A pyramid’s volume is $\frac13Bh$, where $B$ is its base area and $h$ is its perpendicular height, measured straight from the tip to the base. So $1372=\frac13(196)h$. Type $1372\sim\frac13(196)h_1$; Desmos shows $h_1=21$ under PARAMETERS. This vertical height is $21$ inches.

Because the pyramid is right, its tip is directly above the square’s center. From that center to the midpoint of a base edge is half a side, or $14/2=7$ inches. That distance and the $21$-inch vertical height are perpendicular legs. The face’s slant height $\ell$ is across from their right angle, so the Pythagorean theorem gives $\ell^2=21^2+7^2$. A triangular face uses the slant height, not the vertical height, for its area.

Type $21^2+7^2$. Desmos prints $490$, so the slant height squared is $\ell^2=490$.

Take the positive square root because a length can’t be negative:

Factor $490$ to expose a perfect square:

Take $49$ out of the radical:

One face is a triangle with base $14$ and slant height $7\sqrt{10}$, so its area is $\frac12(14)(7\sqrt{10})$. Type $\frac12(14)(7)$; Desmos prints $49$. Keep the $\sqrt{10}$ factor. The area of one triangular face is $49\sqrt{10}$ square inches. Choice A.