Similar cones - volumes and lateral areas Two similar cones have volumes 9$\pi$ and 72$\pi$. If the lateral area of the larger cone is 32$\pi$, what is the lateral area of the smaller cone? I did the following... $\frac {(9\pi)^3} {(32\pi)^2} = \frac {x}{(32\pi)^2}$ resulting in a lateral area of $4\pi$. Is this right?

Since the two cones are similar the ratios of the radius, height and slant height of the larger cone is some multiple of the radius, height and slant height of the smaller cone. Let's call it $k$. This gives $r^{\prime} = kr$, $h^{\prime}=kh$ and $l^{\prime}=kl$ where the primed variables are the dimensions of the larger cone.

We know that $\frac{1}{3}\pi r^2h = 9\pi\Rightarrow r^2h=27$. Likewise, $r'^2h'=216$. Substituting $r' = kr$ and $h' = kh$ gives $(kr)^2kh=216\Rightarrow k^3r^2h=216\Rightarrow k^3(27)=216\Rightarrow k^3=8\Rightarrow k=2$.

Therefore, the larger cone is twice as big as the smaller cone in linear dimensions. We know that the lateral surface of the larger cone is $\pi r'l'=32\pi$ so $\pi (kr)(kl)=32\pi$ from which we get $k^2\pi rl=4\pi rl = 32\pi\Rightarrow \pi rl= 8\pi$.

Therefore, the lateral surface area of the smaller cone is $8\pi$.