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To wit: (see photo)

I don’t see how this can be done for the problem given, as there are two variables (height and radius) and whilst I can work out the volume of both cones with the information given, I’m not sure how to deduce the radius and height of the smaller cone.

Is this a job for simultaneous equations? Is the Pythagorean somehow involved? My efforts at using equation transposition have failed.

enter image description here

duckegg
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1 Answers1

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The volume of a cone is given by

$V = \frac{1}{3} \pi r^2 h $

And if the semi-vertical angle (the angle between the axis of the cone and its surface) is $\theta$ then

$ r = h \tan \theta $

Thus

$V = \frac{1}{3} \pi \tan^2 \theta \hspace{4pt} h^3 $

Now you are given the larger cone volume $V_1$ and smaller cone volume $V_2$, then

$V_1 = \frac{1}{3} \pi \tan^2 \theta \hspace{4pt} h_1^3 $

$V_2 = \frac{1}{3} \pi \tan^2 \theta \hspace{4pt} h_2^3 $

Dividing these two,

$\dfrac{V_1}{V_2} = \left(\dfrac{h_1}{h_2}\right)^3 $

The only unknown here is $h_2$, which according to this last equation is given by

$h_2 = \displaystyle h_1 {\left(\dfrac{V_2}{V_1} \right) }^{\dfrac{1}{3}} $

That is, the smaller cone height is the larger cone height times the cubic root of the ratio of the smaller cone volume $V_2$ to the larger cone volume $V_1$.

Hosam Hajeer
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  • Thanks Hosam, I can follow all of your logic, but I’m just not sure why the index of 3 gets added to h in the equation after your first ‘thus’. – duckegg Oct 14 '21 at 21:47