Question 1153291


total number of diagonals: {{{n(n-3)/2}}} where{{{ n}}} is the number of {{{sides}}}

if the sum of the number of {{{sides}}} and the number of {{{diagonals}}} is {{{66}}}, we have

{{{n+n(n-3)/2=66}}}

{{{(2n+n^2-3n)/2=66}}}

{{{n^2-n=132}}}

{{{n^2-n-132=0}}}...factor completely

{{{n^2+11n-12n-132=0}}}

{{{(n^2+11n)-(12n+132)=0}}}

{{{n(n+11)-12(n+11)=0}}}

{{{(n - 12) (n + 11) = 0}}}

solutions:

{{{n=12}}}
{{{n=-11}}}-> disregard negative solution

 the number of sides is {{{12}}}

answer: B. {{{12}}}