Question 1003640
{{{k}}}= number of sides of the polygon K.
So, {{{k+3}}}= number of sides of polygon L.


A polygon with {{{n}}} sides has {{{n}}} vertices.
Each vertex can be connected to each of the other {{{n-1}}} vertices by a segment.
Two of those segments will be sides, connecting adjacent vertices.
The other {{{n-1-2=n-3}}} segments will be diagonals.
Each of the {{{n}}} vertices is involved in {{{n-3}}} diagonals,
but since each diagonal involves {{{2}}} vertices,
the total number of diagonals for polygon with {{{n}}} sides is
{{{n(n-3)/2}}} .


The total number of diagonals for polygon with {{{k}}} sides is
{{{k(k-3)/2}}} .
The total number of diagonals for polygon with {{{k+3}}} sides is
{{{(k+3)k/2}}} .
The problem says that
{{{(k+3)k/2=36+k(k-3)/2}}} .
We solve for {{{k}}} .
{{{(k+3)k/2=36+k(k-3)/2}}}
{{{(k+3)k=72+k(k-3)}}}
{{{k^2+3k=72+k^2-3k)}}}
{{{3k=72-3k)}}}
{{{6k=72)}}}
{{{k=72/6)}}}
{{{k=12)}}}
{{{k+3=15}}}
Polygon K has {{{highlight(12)}}} sides, and polygon L has {{{highlight(15)}}} sides.