Given a standard football, regular hexagons and also pentagons are drawn on it as presented in the picture. Discover out the number of hexagons and pentagons.

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We can apply Euler attributes to discover out number of Hexagons and also Pentagons ~ above a standard Football.According come Euler Characteristics: because that every surface ar S there exists an creature $$\chi(S)$$

such the whenever a graph G with V vertices and E edges is installed in S so the there are F faces(regions separated by the graph), we have:  $$ V - E + F = \chi(S)$$ For a sphere(shape the football) $$\chi(S)$$= 2.Hence, the equation becomes V – E + F = 2.Now, let variety of pentagons p and number of hexagons H.Number the vertices will be:6 vertices because that each that the hexagons, i.e. 6*H.5 vertices for each of the pentagons, i.e. 5*P.But we have counted each vertex thrice, as soon as for each surrounding polygon, follow the pictureHence, number of vertices, V = (6*H + 5*P)/3.Number of edges will be:6 edges because that each that the hexagons, i.e. 6*H.5 edges for each that the pentagons, i.e. 5*P.But we have actually counted every edge twice, once for each adjacent polygon, monitor the pictureHence, number of edges, E = (6*H + 5*P)/2.

Number of encounters will be:There space H hexagons and P pentagons, each forming a face. Hence, total variety of faces, F = (H + P).So, we have the right to write:  $$ V - E + F = 2 $$ \\ $$ \frac6*H + 5*P3 - \frac6*H + 5*P2 + (H + P) = 2$$ after resolving this equation us will gain P = 12. So, there room 12 Pentagons.Now variety of Hexagons:We have the right to see that each pentagon is surrounded by 5 Hexagons. For this reason there have to 5*P hexagons, yet we have actually counted each hexagon thrice because that each the its 3 nearby pentagons. Hence, number of hexagons = 5*P/3 = 5*12/3 = 20.Hence, there space 20 Hexagons and also 12 Pentagons in a traditional football.
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