Question

Find the sum of exterior angles obtained on producing, in order, the sides of a polygon with :

(i) 7 sides (ii) 10 sides (iii) 250 sides

Answer 1

(i) No. of sides n = 7

Sum of interior & exterior angles $=7\times {180}^o={1260}^o\text{Sum of interior angles}=(n-2)\times {180}^o=(7-2)\times {180}^o={900}^o$

$\therefore \text{Sum of exterior angles}={1260}^o-{900}^o={360}^o$.

(ii) No. of sides n = 10

Sum of interior and exterior angles = $10\times {180}^o={1800}^o\text{But sum of interior angles}=(n-2)\times {180}^o=(10-2)\times {180}^o={1440}^o$

$\therefore \text{Sum of exterior angles}=1800-1440={360}^o$.

(iii) No. of side n = 250

Sum of all interior and exterior angles

$=250\times {180}^o={45000}^o$

But sum of interior angles $=(n-2)\times {180}^o=(250-2)\times {180}^o=248\times {180}^o={44640}^o$

$\therefore \text{Sum of exterior angles}=45000-44640={360}^o$