What is the area of a regular polygon of n sides, each of which is 1 unit?

Dec 20, 2017

The area of regular polygon with $n$ sides and side length
$1$ is ${A}_{p} = \frac{n}{4 \cdot \tan \left(\frac{\pi}{n}\right)}$

Explanation:

The area of regular polygon having sides $n$ and side length

$b$ is A_p= (n *b^2)/(4*tan(pi/n)) ;b =1

$\therefore {A}_{p} = \frac{n}{4 \cdot \tan \left(\frac{\pi}{n}\right)}$

Example : For hexagon of sides 1 ; n=6

${A}_{p} = \frac{6}{4 \cdot \tan \left(\frac{180}{n}\right)} = \frac{6}{4 \cdot \tan 30} \approx 2.60$ sq.unit

The area of regular polygon with $n$ sides and side length

$1$ is ${A}_{p} = \frac{n}{4 \cdot \tan \left(\frac{\pi}{n}\right)}$ [Ans]

Dec 20, 2017

Area of polygon is $\frac{n {b}^{2}}{4} \cot \left({180}^{\circ} / n\right)$ - and if $b = 1$, area is $\frac{n}{4} \cot \left({180}^{\circ} / n\right)$

Explanation:

You are right with the figures. To find the area of the polygon, we should divide it in $n$ isosceles triangles.

Observe that the base is $b$ and height is $h$ (in a polygon this is called apothem). As we draw perpendicular from center of polygon to base, it forms a right angled triangle with base as $\frac{b}{2}$ and height $h$ and as theangle shown $\theta = \frac{{360}^{\circ}}{n}$, the angle $\frac{\theta}{2}$ in right angled triangle is equal to $\frac{{180}^{\circ}}{n}$ and

$\frac{h}{\frac{b}{2}} = \cot \left(\frac{\theta}{2}\right) = \cot \left({180}^{\circ} / n\right)$ and hence

$h = \frac{b}{2} \cot \left({180}^{\circ} / n\right)$

and area of triangle is $\frac{b \times h}{2} = \frac{b}{2} \times \frac{b}{2} \cot \left({180}^{\circ} / n\right)$

or ${b}^{2} / 4 \cot \left({180}^{\circ} / n\right)$

Hence area of polygon is $\frac{n {b}^{2}}{4} \cot \left({180}^{\circ} / n\right)$ - and if $b = 1$, area is $\frac{n}{4} \cot \left({180}^{\circ} / n\right)$