Question

Calculate the number of sides of a regular polygon, if :

(i) its interior angle is five times its exterior angle.

(ii) the ratio between its exterior angle and interior angle is 2 : 7.

(iii) its exterior angle exceeds its interior angle by ${60}^o$.

Answer 1

Let number of sides of a regular polygon = n

(i) Let exterior angle = x

Then interior angle = 5x

$\therefore x+5x={180}^o\Rightarrow 6x={180}^o\Rightarrow 6x={180}^o\Rightarrow x=\frac{{180}^o}{6}={30}^o$

$\therefore$ Number of sides $\left(n\right)=\frac{{360}^o}{30}=12$

(ii) Ratio between exterior angle and interior angle = 2 : 7

Let exterior angle = 2x

Then interior angle = 7x

$\therefore 2x+7x={180}^o\Rightarrow 9x={180}^o\Rightarrow x=\frac{{180}^o}{9}={20}^o$

$\therefore Ext.angle=2x=2\times {20}^o={40}^o$

$\therefore$ No. of sides $=\frac{{360}^o}{40}=9$

(iii) Let interior angle = x

Then exterior angle = x + 60

$\therefore x+x+{60}^o={180}^o\Rightarrow 2x={180}^o-{60}^o={120}^o\Rightarrow x=\frac{{120}^o}{2}={60}^o$

$\therefore$ Exterior angle $={60}^o+{60}^o={120}^o$

$\therefore$ Number of sides $=\frac{{360}^o}{{120}^o}=3$