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Eulers Formula
    The number of faces, edges and vertices of prism and pyramids are connected by the formula :

                               

where V, F and E stand for the number of  vertex , face and edge. This formula is known as Euler’s formula.    
    Cuboid :                 
    A  cuboid has                                      
    (i)     6 rectangular faces                
    (ii)     12 edges            
    (iii)     8 vertices.                               
      Let  F, E and V denote respectively the number of  faces, edges and vertices of a cuboid.
    Then, F – E + V = 6 – 12 + 8 = 2.  

Cube (Square prism) : 
     A  cube has                                                                    
    (i)     6 square faces    
    (ii)     12 edges
    (iii)     8 vertices
    Here, F = 6, E = 12 and V = 8
    F – E + V = 6 – 12 + 8 = 2.

Triangular Pyramid : A triangular pyramid (Tetrahedron) shown in fig. has :     (i)     4 faces 
    (ii)     6 edges
    (iii)     4 vertices
        Here, F = 4, E = 6 and V = 4
           F – E + V = 4 – 6 + 4 = 2

Illustration 1
        A polyhedron has 20 edges and 10 vertices. How many faces does the polyhedron have ?
    Solution 
        Here, E = 20, V = 10
        V + F – E = 2
       10 + F – 20 = 2
        F = 2 – 10 + 20 = 12
    Hence, the polyhedron has 12 faces. 

Illustration 2
        Using Euler’s formula find the unknown.

Solution    
        We know that using Euler’s formula
            F + V – E = 2
        (a)     F + 6 – 12 = 2 
        or     F – 6 = 2 or F = 8
            Thus number of faces = 8
        (b)     5 + V – 9 = 2
            Thus number of vertices = 6
        (c)     2 + 12 – E = 2 
        or     32 – 2 = E or 30 = E
        Thus number of edges = 30

Illustration 3    
        Can a polyhedron have 10 faces, 20 edges and 15 vertices ?
    Solution
        The given data will be of a polyhedron, if it satisfies the Euler’s formula.
        i.e.,    V + F – E = 2
        Here V = 15, F = 10 and E = 20
        Putting in L.H.S.
            15 + 10 – 20 = 5 RHS
            LHS RHS
        Hence given vertices, faces and edges can’t be that of a polyhedron.

 

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