Question 1086341

An n-sided polygon has 
{{{n(n - 3)/ 2}}} diagonals

if a polygon has {{{665}}} diagonals, we have

{{{n(n - 3)/ 2=665}}}

{{{n(n - 3)=665*2}}}

{{{n^2 - 3n=1330}}}

{{{n^2 - 3n-1330=0}}}...factor

{{{(n - 38) (n + 35) = 0}}}

solutions:

{{{(n - 38) = 0}}}=>{{{n=38}}}
{{{(n + 35) = 0}}}=>{{{n=-35}}}=> disregard negative solution if looking for number of the sides

 

b. Why can’t a polygon have {{{406}}} diagonals? 

{{{n(n - 3)/ 2=406}}}}

{{{n(n - 3)=406*2}}}}

{{{n^2 - 3n=812}}}}

{{{n^2 - 3n-812=0}}}}...can't be factored, so use quadratic formula

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

{{{n = (-(-3) +- sqrt( (-3)^2-4*1*(-812) ))/(2*1) }}}

{{{n = (3 +- sqrt( 9+3248 ))/2 }}}

{{{n = (3 +- sqrt( 3257 ))/2 }}}

{{{n = (3 +- 57.07)/2 }}}

solutions: use positive only

{{{n = (3 + 57.07)/2 }}}

{{{n = 60.07/2 }}}

{{{n = 30.035 }}}=> decimal number cannot be solution to number of sides, it have to be an integer because a polygon has integer number of sides