Find the sum of the interior angles of a regular pentagon.

We know that a pentagon has 5 sides.

Therefore, n = 5

Sum of interior angles $=(n-2)\times {180}^o$

$=(5-2)\times {180}^o$

$=3\times {180}^o$

$={540}^o$

Therefore, the sum of the interior angles of a pentagon is ${540}^o.$