I discovered an obvious thing some days back, but is there a solid proof for it? Prove that there are \(\rm x + 1\) faces if the base of a pyramid has \(\rm x\) sides. I know the concept of it, but just need to know if there's a proof.
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The base of a cube has 4 sides but it has 4+2 faces..
I didn't get it!
Oops, yes, pyramids.
It seems pretty obvious. There is one face attached to each of the base polygon's sides, plus the base itself.
Yeah... but I am asking for a solid proof.
@ParthKohli needs a proof involving serious 'mathematics'
For a pentagonal prism:
Pfft, drawing pictures is serious enough for me.
Oh Lord, a solid proof.
LOL, I get it.
Well, @CliffSedge , your diagram says it all...
I need serious mathematics, sire, serious mathematics.
Euler's formula might come into play if you want to get all fancy about it.
to prove something, you need an equation
And that formula is valid also for a cube!!
Euler's formula is valid for all polyhedra, so it will be applicable to the special case of pyramids.
The number of vertices, V, also equals x+1 (x for the vertices of the base polygon, plus the top vertex of the pyramid).
And the number of edges, E = 2x (x for each side of the base polygon, plus the x edges coming up from its x vertices).
These are all based on definitions of pyramids, so
Simplify and identity is verified.
Or since your task is to show that F=x+1, start with
(x+1)-2x+F=2, and solve for F.
This all seems quite circular, since it is merely restating definitions of pyramids.