Question

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.

Answer 1

there is.

Answer 2

Please.

Answer 3

You mean for pyramids?

Answer 4

The base of a cube has 4 sides but it has 4+2 faces..
I didn't get it!

Answer 5

Oops, yes, pyramids.

Answer 6

It seems pretty obvious. There is one face attached to each of the base polygon's sides, plus the base itself.

Answer 7

Yeah... but I am asking for a solid proof.

Answer 8

@ParthKohli needs a proof involving serious 'mathematics'

Answer 9

For a pentagonal prism:

Answer 10

Pfft, drawing pictures is serious enough for me.

Answer 11

Oh Lord, a solid proof.

Answer 12

LOL, I get it.

Answer 13

Induction?

Answer 14

Well, @CliffSedge , your diagram says it all...

Answer 15

I need serious mathematics, sire, serious mathematics.

Answer 16

Euler's formula might come into play if you want to get all fancy about it.
V-E+F=2

Answer 17

to prove something, you need an equation

Answer 18

And that formula is valid also for a cube!!

Answer 19

Euler's formula is valid for all polyhedra, so it will be applicable to the special case of pyramids.
Here F=x+1

Answer 20

The number of vertices, V, also equals x+1 (x for the vertices of the base polygon, plus the top vertex of the pyramid).

Answer 21

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

V-E+F=2
F=x+1
V=x+1
E=2x
(x+1)-2x+(x+1)=2
Simplify and identity is verified.

Answer 22

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.