Question:
What is the sum of all interior angles of a regular
(i) pentagon
(ii) hexagon
(iii) nonagon
(iv) polygon of 12 sides?
Solution:
Sum of the interior angles of an $n$-sided polygon $=(n-2) \times 180^{\circ}$
(i) For a pentagon:
$n=5$ $\therefore(n-2) \times 180^{\circ}=(5-2) \times 180^{\circ}=3 \times 180^{\circ}=540^{\circ}$
(ii) For a hexagon:
$n=6$ $\therefore(n-2) \times 180^{\circ}=(6-2) \times 180^{\circ}=4 \times 180^{\circ}=720^{\circ}$
(iii) For a nonagon:
$n=9$ $\therefore(n-2) \times 180^{\circ}=(9-2) \times 180^{\circ}=7 \times 180^{\circ}=1260^{\circ}$
(iv) For a polygon of 12 sides:
$n=12
$\therefore(n-2) \times 180^{\circ}=(12-2) \times 180^{\circ}=10 \times 180^{\circ}=1800^{\circ}$