Tetrahedron water level

A container shaped like a rotating cylinder with a base radius of 5 cm is filled with water. If a regular tetrahedron with an edge of 7 cm is immersed in it, how much will the water level in the container rise?

Final Answer:

h =  0.5147 cm

Step-by-step explanation:

$$\begin{gathered} r=5\text{cm} \\ {S}_1=\pi \cdot r^2=3.1416\cdot 5^2\doteq 78.5398{\text{cm}}^2 \\ a=7\text{cm} \\ V=\frac{\sqrt{2}}{12}\cdot a^3=\frac{\sqrt{2}}{12}\cdot 7^3\doteq 40.4229{\text{cm}}^3 \\ V=S_1\cdot h \\ h=\frac{V}{S_1}=\frac{40.4229}{78.5398}=0.5147\text{cm} \end{gathered}$$

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