Define the edge of a tetrahedron and then suppose that this tetrahedron begins to be filled from the base. When half of the volume of the body contains water, what height is the part that is not submerged?

### Development:

We define an edge $$= 1 \ m$$.

The volume of the tetrahedron of $$1 \ m$$ of edge is: $$$V=\dfrac{\sqrt{2}}{12}\cdot a^3=0,12 \ m^3$$$

Now we have to find the edge of the not submerged tetrahedron, with half of the volume: $$$0,06=\dfrac{\sqrt{2}}{12}\cdot a'^3 $$$ $$$a'^3=0,5 \Rightarrow a'=0,79 \ m$$$ Then we have to find the height of a face of the tetrahedron: $$$h=a\cdot\dfrac{\sqrt{3}}{2}=0,68 \ m$$$ And, finally, Pythagoras is applied to find the height of the tetrahedron, analyzing the rectangle triangle with base $$\dfrac{a}{2}$$, height $$h'$$ and hypotenuse $$h$$ $$$\Big(\dfrac{a'}{2}\Big)^2+h'^2=h^2$$$ $$$0,395^2+h'^2=0,68^2$$$ $$$h'=0,54 \ m$$$

### Solution:

edge $$= 1 \ m$$

height $$=0,54 \ m$$