Rectangular Prallelipiped

[Kalcor]

You were given areas of three faces of a rectangular parallelepiped. Your task was to find the sum of lengths of it's sides.
Let a, b and c be the lengths of the sides that have one common vertex. Then the numbers we are given are s1 = ab, s2 = bc ,s3 = ca. It is easy to find the lengths in terms of faces areas: a = sqrt(s1*s3 / s2) , b = sqrt(s1*s2 / s3), c = sqrt(s2*s3 / s1) The answer is 4(a + b + c), because there are four sides that have lengths equal to a, b and c. The complexity is O(1).

I don't get the reasons behind the sqrt's , I am not good with shapes xD. I don't know made it reach this solution under the square root. what I concluded is that we multiply by the areas that share a side and divide by the one that doesn't under the square root, but still.. why LOL what property about this "rectangular parallelepiped" makes it work this way?

[lastchance]

"Rectangular parallelepiped" might as well just be called "cuboid" - the edges are orthogonal (at 90 degrees), but it isn't a "cube" because the sides aren't of equal length.

It's unfortunate that they use symbols s1 for ... area!

Anyway, s1 = a * b (or just ab) as the area of a rectangle. Similarly for the others. Then
s1*s3/s2 = ab.ca/(bc) = a². Take the square root, et voilà! sqrt(s1*s3/s2) = a. Similarly for the others.

Thinking of it as a cuboid, there are 4 sides of length a, 4 sides of length b, 4 sides of length c, so the sum of the sharp edges is 4a+4b+4c = 4(a+b+c).

If it wasn't a "rectangular" parallelepiped then, for example, the area
s1 = ab.sin(theta),
where theta is the angle between the sides. Makes the final formulae a lot less pretty. Fundamentally, it's the fact that theta = 90 degrees in a rectangular parallelepiped that "makes it work that way".

If you want the volume of a general parallelepiped (imagine shearing your rectangular parallelepiped sideways) then it is the classic scalar triple product of side vectors a, b and c:
a dot ( b cross c )