I would like to find a way to calculate the height (h1) and base (r2) of the frustum knowing it has the same volume than the underlying cone and the angles are the same.
I got to a 3rd degree unsolvable polynomial and I'm stuck. Any help is welcome.
Thank you
First we note that, since we have the small gray cone, which is part of the larger cone formed when appending the frustrum, we have that $$ \frac{h+h_1}{r_2} = \frac{h}{r} = 5$$ Next, we relate the volumes as follows: \begin{align} V_{\text{little cone}} &= V_{\text{frustrum}} \\ &= V_{\text{big cone}} - V_{\text{little cone}} \\ 2V_{\text{little cone}} &= V_{\text{big cone}} \\ 2\cdot \frac{1}{3}\pi r^2 h &= \frac{1}{3}\pi r_2^2(h+h_1) \\ 2r^2 h &= r_2^2(h+h_1) \\ 2r^2 (5r) &= r_2^2(5r_2) \\ 2r^3 &= r_2^3 \\ r_2 &= \sqrt[3]{2} r \\ &= \sqrt[3]{2} (100 \text{ m}) \\ &\approx 125.992 \text{ m} \end{align} From the first equation, we have \begin{align} h_1 &= 5r_2 - h \\ &= 5(100\sqrt[3]{2} \text{ m}) - 500 \text{ m} \\ &= 500(\sqrt[3]{2}-1) \text{ m} \\ &\approx 129.960 \text{ m} \end{align}