What is the sum of the interior angles of a regular polygon with nine sides?

Sketch the 9-gon. Pick one vertex, and draw all the diagonals to the other verticies.

You should have 6 diagonals drawn: 3 verticies don't have diagonals. The vertex that you picked can't be connected to itself, nor can it be connected to either of the two sides and still be called a "diagonal"...those are sides, not diagonals!

So, count the number of triangles you have. You have 7 triangles. (If you need help, think of cutting a 4-gon, or a square, into two triangles along a single diagonal.) Look at how each of those angles relates to the 9-gon. If you sum up all the angles of the 7 triangles, you ALSO get the sum of all the angles of the 9-gon. So the sum of the angles is 7 * 180.

This is how the formula (n-2)180 for the sum of interior angles is derived. Don't memorize it...understand it!

If you wanted EACH angle in a REGULAR polygon, you'd divide that expression by n.

OK, the sum of each interior angle and it's exterior angle of a regular polygon is 180 degrees. And the sum of the exterior angles is always = 360 degrees. So divide 360 by 9 to get the exterior angle and subtract from 180 to get the interior angle, which would be 140. Now multiply by 9 to get your answer.