# SOLUTION: 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!!! . Base of polyhedral Vertices Faces

Algebra ->  Algebra  -> Rectangles -> SOLUTION: 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!!! . Base of polyhedral Vertices Faces      Log On

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 Geometry: Rectangles Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Rectangles 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!!! . 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 ?