Question 367070: Hi, I'm hoping someone can please help me! I'm having issues trying to figure out this problem. Your help is greatly appreciated! Thank you!!!
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Base of polyhedral Vertices Faces Edges
Triangle 4vertices 4faces 6edges
Square ____vertices ____faces 8edges
Pentagon ____vertices 6faces ____edges
Hexagon 7vertices ____faces ____edges
What type of polyhedral is represented in the table?
If n represents the number of sides of the polyhedra base, then write an equation for the number of vertices, faces, and edges of the polyhedral.
Answer by Jk22(389)   (Show Source):
You can put this solution on YOUR website! Supposing the problem were to build a polyhedron with the same regular polygon as faces :
Taking Euler's formula for polyhedron, we have :
V - E + F = 2 (vertex minus edge + face number equals 2)
suppose the faces were the same base polygon with n side (n vertices)
then :
n*F/2 = E (each side of the F polygons (with n sides) are shared by 2 faces)
n*F/3 = V (each vertex of the n*F is shared by 3 faces)
the equation becomes : n*F/3 - n*F/2 + F = 2
or (6-n)F = 12
F = 12/(6-n), E = 6n/(6-n), V = 4n/(6-n)
triangle : n = 3 : V = 12/3 = 4, E = 18/3 = 6, F = 12/3 = 4
square : n = 4 : V = 16/2 = 8, E = 24/2 = 12, F = 12/2 = 6
pentagon : n = 5 : V = 20, E = 30, F = 12
hexagon : n = 6, all are infinite, this covers the plane.
if the vertex is shared by m faces : V = n*F/m
n*F/m - n*F/2 + F = 2
(2m + 2n - nm)F = 4m
F = 4m/(m(2-n)+2n)
if n=3 (base triangle), F = 4m/(6-m),
m-----3----4-----5
4m----12---16----20
6-m---3----2-----1
-----------------
F-----4----8-----20
tetrahedron, octahedron and icosahedron,
if n=4 (base square), F = 4m/(8-2m)
m=3, F = 12/2 = 6 (cube) (only 3 faces per vertex possible)
n=5 (base pentagon), F = 4m/(10-3m), => m=3 dodecahedron
other solutions : the number of faces per vertex are mixed (like complementary fullerenes) : suppose base as a triangle
suppose n1 and n2 are the possible number of faces per vertex, with 3F/k vertices n2-connected, and 3F(1-1/k) n1-connected
F(n2(6-n1) + 6/k*(n1-n2)) = 4n1n2
for n1=5, n2=3, : F=12, k=6 : 10 vertices 5-connected, 2 vertices 3-connected
for n1=3, n2=5, F=20, k=2, 10 vertices 3-connected, and 10 5-connected
if the polyhedron is made out of 2 faces' type, there are also other cases :
polyhedra called fullerenes
(hexagons and other polygons, or other mix of polygons).
the formula becomes : n1*(6-s1) + n2*(6-s2) = 12, where ni are the numbers of regular polygons with si sides.
This gives solution like :
heptagons (2,3,4,6) and pentagons (14, 15, 16, 18)
4 heptagons and 8 squares
or heptagons (6) and triangles (6) should be able to build a polyhedron ?
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