Find the sum of the measures of the interior angles of each of the following polygons: a pentagon, an octagon, a dodecagon, a 40-gon, a 52-gon, and a 100-gon.
The sum of the measures of the interior angles of a polygon with $n$ sides is $180(n-2)$ degrees. Thus, the sums of the measures of the interior angles of the given polygons are $180(5-2) = \boxed{540}$, $180(8-2) = \boxed{1080}$, $180(12-2) = \boxed{1800}$, $180(40-2) = \boxed{6840}$, $180(52-2) = \boxed{9000}$, and $180(100-2) = \boxed{17640}$ degrees, respectively.
