Question 96·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

Cone A has volume $100\pi$ and radius $5$, so put those values into the given cone volume formula:

Here $h_A$ means cone A’s perpendicular height, measured straight down from the tip.

To solve that equation in Desmos, replace $=$ with $\sim$ and use $h_1$ for the unknown height, rather than creating a slider. Type $100\pi\sim\frac{1}{3}\pi(5)^2h_1$. Desmos shows $h_1=12$ under PARAMETERS, so cone A’s perpendicular height is $12$ cm.

The slant height runs from the tip along the cone’s side. Use the given formula with radius $5$ and perpendicular height $12$:

These make a $5$-$12$-$13$ right triangle. Use $13$, not $12$, for the curved surface.

The table calls cone A’s surface area $k\pi$. Its base contributes $\pi r^2$, and its curved side contributes $\pi r\ell$, so type $k=5^2+5(13)$ in Desmos. It gives $k=90$, meaning cone A’s surface area is $90\pi$.

To go from A to B, divide B’s volume by A’s. The $\pi$ factors cancel. Type $12500/100$ in Desmos; it gives $125$, so cone B’s volume is $125$ times cone A’s volume.

Let $s$ be the linear scale factor: every matching length in B is $s$ times the length in A. Volume uses three lengths, so its factor is $s^3$:

Take the positive cube root because lengths are positive. Type $s_1^3\sim125$ in Desmos; it shows $s_1=5$ under PARAMETERS, so:

Each term of surface area uses two lengths: $r^2$ or $r\ell$. Volume scales by $s^3$, but surface area scales by $s^2$. With $s=5$, type $j=k(5^2)$ in Desmos. It gives $j=2250$, the coefficient of $\pi$ for cone B.

Type $j-k$ in Desmos; it gives $2160$. The question asks for the difference between the coefficients, not the difference between the surface areas, so the value of $j-k$ is $2160$, without $\pi$. Choice C.