Question

In each of the following polygons find in degrees the sum of the interior angles and the sum of exterior angles.
(i) Hexangon (ii) Octagon (iii) Pentagon (iv) Nanogon (v) Decagon.

Answer 1

(i) Hexagon

Numbers of sides, $n=6$

Sum of the interior angles $=(2n–4){90}^0$

$=\left(\right(2\times 6)–4)\times {90}^0$

$=(12–4){90}^0$

$=8\times {90}^0$

$={720}^0$

Sum of the exterior angles $={360}^0$.

(ii) Octagon

Numbers of sides, $n=8$

Sum of the interior angles $=(2n–4){90}^0$

$=\left(\right(2\times 8)–4)\times {90}^0$

$=(16–4){90}^0$

$=12\times {90}^0$

$={1080}^0$

Sum of the exterior angles $={360}^0$.

(iii) Pentagon

Numbers of sides, $n=5$

Sum of the interior angles $=(2n–4){90}^0$

$=\left(\right(2\times 5)–4)\times {90}^0$

$=(10–4){90}^0$

$=6\times {90}^0$

$={540}^0$

Sum of the exterior angles $={360}^0$.

(iv) Nonagon

Numbers of sides, $n=9$

Sum of the interior angles $=(2n–4){90}^0$

$=\left(\right(2\times 9)–4)\times {90}^0$

$=(18–4){90}^0$

$=14\times {90}^0$

$={1260}^0$

Sum of the exterior angles $={360}^0$.

(v) Decagon

Numbers of sides, $n=10$

Sum of the interior angles $=(2n–4){90}^0$

$=\left(\right(2\times 10)–4)\times {90}^0$

$=(20–4){90}^0$

$=16\times {90}^0$

$={1440}^0$

Sum of the exterior angles $={360}^0$.