Question 750743
Let {{{n}}} be the number of sides = number of vertices.

Then {{{each}}} {{{vertex}}} can be connected to {{{n-3}}} other vertices, giving {{{n(n -3)}}} connections. However, this counts each vertex pair connection twice, so there are:

{{{n(n - 3) / 2}}} diagonals.

Thus if there are {{{65}}} diagonals:

{{{n(n - 3) / 2 = 65}}}

{{{n^2 - 3n = 65*2}}}

{{{n^2 - 3n - 130 = 0}}}...write {{{-3n}}} as {{{10n-13n}}}

{{{n^2+10n-13n-130=0}}}...group

{{{(n^2+10n)-(13n+130)=0}}}

{{{n(n+10)-13(n+10)=0}}}

{{{(n-13)(n+10) = 0}}}

so {{{n = 13}}} or {{{n=-10}}}

As a polygon cannot have a negative number of sides, {{{n = 13}}}.




 if there are {{{80}}} diagonals:

{{{n(n - 3) / 2 = 80}}}

{{{n^2 - 3n = 80*2}}}

{{{n^2 - 3n - 160 = 0}}}...use quadratic formula


 {{{n = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


 {{{n = (-(-3) +- sqrt((-3)^2-4*1*(-160) ))/(2*1) }}}

 {{{n = (3 +- sqrt(9+640) ))/2 }}}

{{{n = (3 +- sqrt(649) ))/2 }}}

{{{n = (3 +- 25.48 ))/2 }}}

solutions:

{{{n = (3 + 25.48 ))/2 }}}

{{{n = 28.48/2 }}}

{{{n = 28.48/2 }}}

{{{n = 14.24 }}}

so, since {{{n = 14.24}}} and {{{n}}} is equal to the number of sides, means that a polygon cannot have a number of sides as decimal number

there is {{{no}}} a polygon with {{{80}}} diagonals