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Find the number of sides of a regular polygon whose each exterior angle is half of it's interior angle
Given :
The exterior angle of a regular polygon is half of its interior angle.
To do :
We have to find the number of sides of the polygon.
Solution :
Let the number of sides of the regular polygon be 'n'.
The exterior angle of a regular polygon with n sides $= \frac{360}{n}$
The interior angle of a regular polygon with n sides $=180 - \frac{360}{n}$
Here, the exterior angle is half of its interior angle.
$\frac{360}{n} = \frac{1}{2}(180 - \frac{360}{n})$
$\frac{360\times 2}{n} = 180 - \frac{360}{n}$
$\frac{360\times 2}{n} = \frac{180 n - 360}{n}$
$360\times 2 =180 n - 360 $ [$n$ on both sides get cancelled]
$360\times 2 + 360=180 n $
Take 360 as common in LHS,
$360(2 + 1) = 180 n $
$360 \times 3 = 180 n$
Rewrite,
$180 n = 360 \times 3$
$n = \frac{360 \times 3}{180}$
$n = 2 \times 3$ $[\frac{360}{180} = 2]$
$n = 6$
Therefore, the number of sides of the regular polygon is 6.
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