Problem:
A wooden cone of altitude 10cm is to be cut into two parts of equal weight. How far from the vertex should the cut parallel to the base be made?
Solution:
to have two parts of equal weight, the approach would be of equal volume
by weight = by volume
r = radius of original cone
h = height of original cone
R = radius of cutout cone
H = height of cutout cone
Volume of cone
[tex]\[V = \frac{1}{3}\pi {r^2}h\][/tex]
Using similar triangle or similarity of the cones, the ratio will be
[tex]\[\frac{h}{r} = \frac{H}{R}\][/tex]
[tex]\[r = \frac{{Rh}}{H}\][/tex]
Half volume of original cone (because of two parts of equal weight) = Volume of cutout cone
Half volume of original cone = Volume of cutout cone
[tex]\[\begin{array}{l}\frac{{\frac{1}{3}\pi {r^2}h}}{2} = \frac{1}{3}\pi {R^2}H\\\\\frac{{{r^2}h}}{2} = {R^2}H\end{array}\][/tex]
[tex]\[\begin{array}{l}{(\frac{{Rh}}{H})^2}\frac{h}{2} = {R^2}H\\\\\frac{{{R^2}{h^3}}}{{2{H^2}}} = {R^2}H\\\\{H^3} = \frac{{{h^3}}}{2}\end{array}\][/tex]
[tex]\[\begin{array}{l}{H^3} = \frac{{{{10}^3}}}{2}\\\\{H^3} = \frac{{1000}}{2}\\\\{H^3} = 500\\\\H = \sqrt[3]{{500}}\\\\H = 7.937005259841cm\end{array}\][/tex]
Answer:
The cut should be 7.937005259841 cm from the vertex
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