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

Question

Claire Akin · Accepted Answer

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

Savannah Aguilar · Answer

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

Sadie Adams · Answer

undefined The area of a regular polygon with $ n $ sides, each of which is 1 unit, can be calculated using the formula:

$$ \text{Area} = \frac{1}{4} n \tan\left(\frac{\pi}{n}\right) $$

Where:
- $ n $ is the number of sides of the polygon.
- $ \tan\left(\frac{\pi}{n}\right) $ is the tangent of the angle formed by connecting the center of the polygon with any of its vertices.

Substituting $ 1 $ for each side length, the formula becomes:

$$ \text{Area} = \frac{1}{4} n \tan\left(\frac{\pi}{n}\right) $$

This formula gives the area of the regular polygon with $ n $ sides, each of which is 1 unit.