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Question 1182459: If the bottom face of a pyramid has n sides, how many faces, edges, and vertices will it
have?
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
Draw a picture, at least in your mind....
Vertices: n sides on the base, so there are n vertices on the base; then just one more -- the peak of the pyramid.
ANSWER: (n+1) vertices
Edges: n sides of the base, each of which is an edge of the pyramid; then one edge from each of the n base vertices to the peak.
ANSWER: n+n=2n edges
Faces: the base, plus n triangular faces.
ANSWER: (n+1) faces.
Note those numbers satisfy the Euler formula:
edges = faces plus vertices minus 2
2n = (n+1)+(n+1)-2
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answers:
n+1 faces
2n edges
n+1 vertices
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Explanation:
The pyramid has an n-sided polygon as the base, where n is some positive whole number larger than 2.
Let's say n = 5 and we have a pentagonal pyramid
There are n = 5 lateral sides (each lateral side a triangle), plus we have the base face as well on the bottom. That gives n+1 = 5+1 = 6 faces total
Going back to the general case, we would simply have n+1 faces.
n lateral faces, plus that 1 base face.
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Consider a triangular pyramid (n = 3) having 3 lateral edges that all meet at the apex or highest point of the pyramid.
Along the bottom there are n = 3 more edges.
Overall, this triangular pyramid has n+n = 3+3 = 6 edges.
In general, we have n+n = 2n edges.
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Along the bottom there are n vertices. At the very top is another vertex. So we have n+1 vertices overall.
We can use Euler's polyhedron formula
V - E + F = 2
to help confirm the answers above
V = number of vertices = n+1
E = number of edges = 2n
F = number of faces = n+1
So,
V - E + F = 2
n+1 - 2n + n+1 = 2
2 = 2
showing that everything works out
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