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 =  3.464 cm

Solution:

r=6 cm s=2⋅ r=2⋅ 6=12 cm h=s2−r2=122−62≐6 3 cm≐10.3923 cm  V=13⋅ π⋅ r2⋅ h=13⋅ 3.1416⋅ 62⋅ 10.3923≐391.7807 cm3 V1=13⋅ V=13⋅ 391.7807≐130.5936 cm3 V2=V−V1=391.7807−130.5936≐261.1871 cm3  V1=13⋅ π⋅ r2⋅ h2  h2=3⋅ V1π⋅ r2=3⋅ 130.59363.1416⋅ 62=2 3=3.464 cmr=6 \ \text{cm} \ \\ s=2 \cdot \ r=2 \cdot \ 6=12 \ \text{cm} \ \\ h=\sqrt{ s^2 - r^2 }=\sqrt{ 12^2 - 6^2 } \doteq 6 \ \sqrt{ 3 } \ \text{cm} \doteq 10.3923 \ \text{cm} \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ \pi \cdot \ r^2 \cdot \ h=\dfrac{ 1 }{ 3 } \cdot \ 3.1416 \cdot \ 6^2 \cdot \ 10.3923 \doteq 391.7807 \ \text{cm}^3 \ \\ V_{1}=\dfrac{ 1 }{ 3 } \cdot \ V=\dfrac{ 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}=\dfrac{ 1 }{ 3 } \cdot \ \pi \cdot \ r^2 \cdot \ h_{2} \ \\ \ \\ h_{2}=\dfrac{ 3 \cdot \ V_{1} }{ \pi \cdot \ r^2 }=\dfrac{ 3 \cdot \ 130.5936 }{ 3.1416 \cdot \ 6^2 }=2 \ \sqrt{ 3 }=3.464 \ \text{cm}

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