<PRE><B>Question 41213</B>
If diagonals are the lines that connect minus the sides to make the 21-gon:
I have thought about polygons like these, and my idea can make this easy:
(1) + (2) + (3) + .... for (s - 1) amount of times where {{{s}}} is the number of sides
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ....
{{{SUM = (n/2)(a1 + an)}}}
{{{SUM = (20/2)(1 + 20)}}}
{{{SUM = (10)(21)}}}
{{{SUM = 210}}}
minus the sides that connect:
{{{210 - 21 = x}}}
{{{189 = x}}}</PRE>