# Finding number of sides of a polygon with an unknown

A regular polygon has $n$ sides . When the number of sides is doubled, each interior angle increases by $20^{\circ}$. Find $n$.

My workings till I got stuck

$1$ int. angle of $n$ sides $=180^{\circ}n-360^{\circ}/n$

$1$ int. angle of $2n$ sides $=200^{\circ}n-360^{\circ}/n$

$1$ ext. angle of $2n$ sides $=-20^{\circ}n+360^{\circ}/n$

$n=360$ divide $1$ ext. angle ...

I've done it till $-20n+360=360$

Then I got stuck. Can I get help? Thanks in advance!

## 3 Answers

The total of the angles in an $n$-gon is $180(n-2)$, so each angle is

$$a(n) = \frac{180(n-2)}{n}$$

with the angle measured in degrees.

So you want to find $n$ where $a(2n) = a(n) + 20$. This is the same as solving

$$\frac{180(2n-2)}{2n} = \frac{180(n-2)}{n} + 20.$$

Multiplying through by $n$ gives

$$180(n-1) = 180(n-2) + 20n,$$

so $n=9$.

Angle sum of $n$-sided polygon $=(n-2) \times 180^{\circ}$

Now, \begin{align*} \frac{(2n-2)}{2n} \times 180^{\circ}- \frac{(n-2)}{n} \times 180^{\circ} &= 20^{\circ} \\ [(n-1)-(n-2)] \times 180^{\circ} &= n\times 20^{\circ} \\ 180 &= 20n \\ n &= 9 \end{align*}

The sum of the exterior angles is $360^\circ$, so each exterior angle on the original polygon is $(360/n)^\circ$.

Doubling the number of sides - now $2n$ - decreases the exterior angle by $20^\circ$. So:

\begin{align} 360/n -20 &= 360/2n \\ 360 -20n &=180\\ 20n &=180\\ n &=9\\ \end{align}