Question

What is the sum of all interior angles of a pentagon? $\left[1\right]$
(a) ${360}^{\circ }$ (b) ${540}^{\circ }$ (c) ${720}^{\circ }$ (d) ${900}^{\circ }$

Answer 1

The sum of the interior angles of a polygon of n sides is $(n-2)\times {180}^{\circ }$
Then, the sum of the interior angles of pentagon is $(5-2)\times {180}^{\circ }={540}^{\circ }$