Question 29·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

In a right square pyramid, the tip sits directly above the center of the base. The distance from that center to the midpoint of an edge is half the side length, so it is $6/2=3$ centimeters.

The $3$-centimeter distance along the base and the given $4$-centimeter vertical height meet at a right angle. They form a $3$-$4$-$5$ triangle, so the face’s slant height, its sloped height from tip to edge, is $5$ centimeters. That isn’t the vertical height used for volume.

The square base has area $6^2$, and each triangular face has area $\frac12(6)(5)$. Type $a=6^2+4\left(\frac12\right)(6)(5)$ in Desmos; it shows $a=96$. Because the problem says including its base, $96$ square centimeters is pyramid A’s total surface area.

Let $k$ be the linear scale factor from A to B: every matching length in B is $k$ times as long. Each face’s area uses two lengths, so every face, and their total, is multiplied by $k^2$. Type $q=480/a$ in Desmos; it shows $q=5$, the area ratio. So $k^2=\frac{S_B}{S_A}=\frac{480}{96}=5$.

Take the positive square root, since lengths are positive: $k=\sqrt{5}$. Type $k=\sqrt{q}$ in Desmos; it shows about $2.236$, confirming that B’s lengths are larger than A’s.

A pyramid’s volume is one-third of its base area times its perpendicular height. Type $u=\frac13(6^2)(4)$ in Desmos; it shows $u=48$. So pyramid A has volume $48$ cubic centimeters. Use the given height $4$, not the face slant height $5$.

Volume uses three scaled lengths, so type $v=uk^3$ in Desmos. It shows about $536.656$ cubic centimeters. Keep the answer exact by substituting the values found above: $V_B=48(\sqrt5)^3$. Use $(\sqrt5)^2=5$ to rewrite the cube: $V_B=48(5\sqrt5)$. Multiply: $V_B=240\sqrt5$. Pyramid B’s volume is $240\sqrt5$ cubic centimeters. Choice A.