Find the sum of the measures of the interior angles of each of the following convex polygons: a pentagon, an octagon, a dodecagon, a 40-gon, a 52-gon, and a 100-gon.

Answer 1

The formula for the sum of the measures of the interior angles of a polygon with $n$ sides is $(n-2)\cdot180^\circ$.

For a pentagon, $n=5$, so the sum of the interior angles is $(5-2)\cdot180^\circ=540^\circ$.

For an octagon, $n=8$, so the sum of the interior angles is $(8-2)\cdot180^\circ=1080^\circ$.

For a dodecagon, $n=12$, so the sum of the interior angles is $(12-2)\cdot180^\circ=\boxed{1800^\circ}$.

For a 40-gon, $n=40$, so the sum of the interior angles is $(40-2)\cdot180^\circ=6840^\circ$.

For a 52-gon, $n=52$, so the sum of the interior angles is $(52-2)\cdot180^\circ=9000^\circ$.

For a 100-gon, $n=100$, so the sum of the interior angles is $(100-2)\cdot180^\circ=16200^\circ$.