Icosagon

If a polygon has n sides, then sum of interior angles of polygon is (n-2)*180
You can check the proof at www.khanacademy.com

There are 2 ways to solve this problem.

Firstly, the more direct method is to use the formula where n is the number of edges or angles in the polygon (n-2) 180 = sum of interior angles, which gives: (20-2) 180 = 18*180 = 3240.

Also, the sum of exterior angles in a polygon is 360. Thus, by dividing that number by the number of edges (20), you can find the size of each exterior angle to be 18. As this angle is on a straight line with the interior angle, the interior angle is then 180-18 = 162. As this only give one individual interior angle, multiply by 20 (the number of angles) to give 3240.

There are 2 ways of solving this problem

Method 1

It is simple we can use this formula

$n-2\times180$

and it is equal to $18\times180$

which is equal to

$=3240$

Method 2

We can multiply $162$ which is the size of each interior angle of an icosagon and then multiply it by $20$ which is the number of sides.

This will also equal to

$=3240$

Sum of interior angles of a polygon $S_n=(n-2)\times180$

So. $S_{20}=(20-2)\times180=\boxed{3240}$