Question 173·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

A square’s diagonal is the hypotenuse of a $45$-$45$-$90$ triangle, so it equals the side length $s$ times $\sqrt{2}$. The given diagonal makes $s\sqrt{2}=30\sqrt{2}$. Divide by $\sqrt{2}$:

The base is a square with side $30$, so its area is

square centimeters.

A pyramid’s volume is one-third of its base area times its perpendicular height $H$, the distance straight down from its tip to the base. Put the given volume and the base area into $V=\frac13BH$:

Type $6000\sim\frac13(900)x_1$ in Desmos, using $x_1$ for $H$. The $\sim$ tells Desmos to find the value that makes the two sides equal. Under PARAMETERS, it gives $x_1=20$, so the perpendicular height is $20$ centimeters.

Because the pyramid is right, its tip is above the square’s center. From that center to the midpoint of a $30$-centimeter edge is $30/2=15$ centimeters. The face’s slant height runs from the tip to that midpoint, not straight down to the base. So $15$ and the perpendicular height $20$ are the legs of a right triangle whose hypotenuse is the slant height.

Use the $3$-$4$-$5$ right-triangle pattern scaled by $5$: legs of $15$ and $20$ have a hypotenuse of $25$. So the face’s slant height is $25$ centimeters.

For one triangular face, use half its base times its own height. Type $\frac12(30)(25)$ on a new Desmos line; it displays $375$. The area of one triangular face is $375$ square centimeters. Choice B.