Question

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

Answer 1

The area of regular polygon with `n` sides and side length
`1` is `A_p=n/(4*tan(pi/n))`

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`

`:. A_p= n/(4*tan(pi/n))`

Example : For hexagon of sides `1 ; n=6`

`A_p= 6/(4*tan(180/n))=6/(4*tan30)~~2.60` sq.unit

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

`1` is `A_p=n/(4*tan(pi/n))` [Ans]

Answer 2

Area of polygon is `(nb^2)/4cot(180^@/n)` - and if `b=1`, area is `n/4cot(180^@/n)`

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 `b/2` and height `h` and as theangle shown `theta=(360^@)/n`, the angle `theta/2` in right angled triangle is equal to `(180^@)/n` and

`h/(b/2)=cot(theta/2)=cot(180^@/n)` and hence

`h=b/2cot(180^@/n)`

and area of triangle is `(bxxh)/2=b/2xxb/2cot(180^@/n)`

or `b^2/4cot(180^@/n)`

Hence area of polygon is `(nb^2)/4cot(180^@/n)` - and if `b=1`, area is `n/4cot(180^@/n)`