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?
$\endgroup$2 Answers
$\begingroup$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 $8\pi$.
$\endgroup$ $\begingroup$If you scale lengths by a factor $f$, then areas scale by $f^2$ and the volume scales by $f^3$. So the factor between the larger and the smaller cone is
$$ f = \sqrt[3]{\frac{72\pi}{9\pi}} = \sqrt[3]8 = 2 $$
Now you can scale the area with that. Going from larger to smaller, you have to divide by $f^2$.
$$ \frac{32\pi}{f^2} = \frac{32\pi}4 = 8\pi $$
So your result appears to be incorrect. The equation from which you derive it looks like a copy and paste error: both sides are completely equal for $x=(9\pi)^3$, which has nothing to do with the answer you gave.
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