Equilateral cone

We pour so much water into a container that has the shape of an equilateral cone, the base of which has a radius r = 6 cm, that one-third of the volume of the cone is filled. How high will the water reach if we turn the cone upside down?

Correct result:

h₂ =  3.4641 cm

Solution:

$$\begin{gathered} r=6\text{ cm } \\ s=2\cdot r=2\cdot 6=12\text{ cm } \\ h=\sqrt{s^2-r^2}=\sqrt{12^2-6^2}=6\sqrt{3}\text{ cm}\doteq 10.3923\text{ cm } \\ V={\displaystyle \frac{1}{3}}\cdot \pi \cdot r^2\cdot h={\displaystyle \frac{1}{3}}\cdot 3.1416\cdot 6^2\cdot 10.3923\doteq 391.7807{\text{cm}}^3 \\ {V}_1={\displaystyle \frac{1}{3}}\cdot V={\displaystyle \frac{1}{3}}\cdot 391.7807\doteq 130.5936{\text{cm}}^3 \\ {V}_2=V-V_1=391.7807-130.5936\doteq 261.1871{\text{cm}}^3 \\ {V}_1={\displaystyle \frac{1}{3}}\cdot \pi \cdot r^2\cdot h_2 \\ {h}_2={\displaystyle \frac{3\cdot V_1}{\pi \cdot r^2}}={\displaystyle \frac{3\cdot 130.5936}{3.1416\cdot 6^2}}=2\sqrt{3}=3.4641\text{ cm} \end{gathered}$$

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