Problems from Tetrahedron: Surface area and volume

Development:

We define an edge $=1m$.

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

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

Solution:

edge $=1m$

height $=0,54m$

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