The octahedron is a polyhedron of eight faces, regular when all the faces are equilateral triangles.
The following expressions are used to find the area and volume of the octahedron $$$A=2 \cdot \sqrt{3} \cdot a^2 \\ V=\dfrac{\sqrt{2}}{3} a^3$$$
a) Define the dimensions of an octahedron (edge $$a$$).
b) Imagine that this octahedron is an iceberg, with the square in the middle (the base of two pyramids that form it) parallel to the sea. A small square pyramid of height $$h'$$ remains out of the water, and the entire octahedron has height $$2 \cdot h$$ ($$h$$ is the height of each of two pyramids that compose the octahedron). Find the height of half an octahedron, $$h$$.
c) Decide a reasonable percentage of $$h'$$ according to $$h$$ for an iceberg, and find the parameter $$a'$$.
d) Which percentage of the volume of the octahedron reminds out of the water?
e) What is the area of the completely submerged faces of the iceberg (four low ones)?
Let's see the solution.
a) We define an edge as $$a = 10 \ m$$, which is reasonable for being an iceberg.
b) To find $$h$$, we must first find the height of the triangle of each of the side faces $$Ap$$. $$$10^2= Ap^2+\Big(\dfrac{10}{2}\Big)^2 \\ Ap= 5\sqrt{3}$$$
Using the Pythagoras’s theorem, with the triangle which sides are $$h$$ and $$\dfrac{a}{2}$$, and which hypotenuse is $$Ap$$, the height of half an octahedron ,$$h$$, is: $$$Ap^2= h^2+\Big(\dfrac{a}{2}\Big)^2 \\(5\sqrt{3})^2=h^2+5^2 \\ h= \sqrt{75-25}=5\sqrt{2}= 7,07 \ m $$$
c) We define $$h' =0,2 \cdot h=1,41 \ m$$, which is reasonable if we think that the main part of an iceberg is always submerged.
We use the Thales theorem to find the edge of the pyramid that reminds out of the water: $$$\dfrac{a'}{a}=\dfrac{h'}{h}=0, 2 \\ a'=2 \ m$$$
d) The entire volume of the octahedron is calculated with the following formula: $$$V=\dfrac {\sqrt{2}}{3} \cdot a^3$$$ where $$a = 10 \ m$$. $$$V_{octo}=471,4 \ m^3$$$
Now, we calcule the volume of the pyramid that remains out of the water: $$$V_{out \ of \ the \ water}=\dfrac{a'^2 \cdot h'}{3}= 1,88 \ m^3$$$ The percentage of the volume out of the water is $$$\dfrac{1,88}{471,4}\cdot 100= 0,39 \%$$$
e) Now we just need to calculate the area of the octahedron and divide it by $$2$$: $$$\dfrac{A_{octo}}{2}=\dfrac{1}{2}(2 \sqrt{3} \cdot a^2) $$$ where $$a=10$$ $$$\dfrac{A_{octo}}{2}=173,2 \ m^2$$$