Question 652123
The way I understand geometry, the answer is {{{highlight(NONE)}}} .
I am betting there is a mix-up in the wording of the problem.
 
The way I understand geometry,
a diagonal of a polygon is a segment that connects two vertices of the polygon, but is not a side of the polygon.
If a polygon has {{{n}}} sides, it will have n vertices.
Each vertex will be connected to {{{2}}} other vertices by sides,
and can be connected to the remaining {{{n-3}}} other vertices by a diagonal.
If we multiply the {{{n-3}}} diagonals coming out of each vertex times {{{n}}} , the number of vertices,
we get {{{n(n-3)}}} ,
but we are counting each diagonal twice.
The polygon with {{{n}}} sides has {{{d=n(n-3)/2}}} diagonals.
 
To find what polygon has a number of sides that is five times the number of diagonals, we can set up an equation
{{{n=5*(n(n-3)/2)}}}
or we can make a table comparing {{{n}}} and {{{d}}} .

For that table:
A triangle has {{{3}}} sides and {{{0}}} diagonals.
A quadrilateral has {{{4}}} sides and {{{2}}} diagonals.
From then on, the number of sides is equal to or less than the number of diagonals.
A pentagon has {{{5}}} sides and {{{5}}} diagonals.
A hexagon has {{{6}}} sides and {{{9}}} diagonals.
 
Either way (equation or table), the answer is {{{highlight(NONE)}}} . 
 
If we wanted to know what polygon has a number of diagonals that is five times the number of sides, our equation would be
{{{5n=n(n-3)/2}}}
{{{5n=n(n-3)/2}}} --> {{{2*5n=2*(n(n-3)/2)}}} --> {{{10n=n(n-3)}}} --> {{{10n=n^2-3n)}}} --> {{{10n+3n=n^2-3n+3n)}}} --> {{{13n=n^2)}}}
Since {{{n=0}}} does not make a polygon, we know {{{n}}} is not {{{0}}} .
{{{13n=n^2)}}} --> {{{13n/n=n^2/n)}}} --> {{{13=n)}}} or {{{highlight(n=13)}}} .