Question

1]a polyhedron has 5 more edges than vertices.how many faces does it have ?
2]how many faces does a polyhedron have that has same no. faces as vertices and 2 more edged than vertices?

Answer 1

Dear student,

$$\begin{gathered} Weknowthatinapolyhedron \\ IfF=faces,V=verticesandE\hspace{0.17em}=edges \\ Then, \\ F+\hspace{0.17em}V\hspace{0.17em}-E=2 \\ Now, \\ \left(1\right)Accordingtothequestion, \\ E=5+\hspace{0.17em}V \\ Then, \\ F+\hspace{0.17em}V-E=2 \\ \Rightarrow F\hspace{0.17em}+\hspace{0.17em}V-5-V=2 \\ \Rightarrow F\hspace{0.17em}-5=2 \\ \Rightarrow F\hspace{0.17em}=7 \\ So,thegivenpolyhedronwillhave7faces. \\ \left(2\right)Accordingtothequestion, \\ E\hspace{0.17em}=V+2---\left(i\right),and \\ F\hspace{0.17em}=V----\left(ii\right) \\ PutingthevalueofVfrom\left(i\right)in\left(ii\right)weget: \\ F=V\hspace{0.17em}=E\hspace{0.17em}-2---\left(iii\right) \\ So, \\ F\hspace{0.17em}+\hspace{0.17em}V-\hspace{0.17em}E=2 \\ \Rightarrow E-2+\hspace{0.17em}E-2-E=2 \\ \Rightarrow E-4=2 \\ \Rightarrow E=6 \\ PuttingthevalueofEin\left(iii\right).Weget, \\ F=\hspace{0.17em}E-2=6-2=4 \\ \therefore Thegivenpolyhedronwillhave4faces. \end{gathered}$$
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