If-you-have-a-regular-n-sided-polygon-is-there-a-method-to-calculate-the-area-from-one-corner-to-another-That-is-if-we-start-at-a-corner-corner-1-and-draw-a-line-to-corner-x-what-is-the-area-

Question Number 5515 by FilupSmith last updated on 17/May/16

If you have a regular n - sided polygon,

is there a method to calculate the area

from one corner to another ?

That is, if we start at a corner (corner 1),

and draw a line to corner x, what is the area ?

See image in comment for visual representation .

Commented by FilupSmith last updated on 17/May/16

Answered by Rasheed Soomro last updated on 17/May/16

Commented by Rasheed Soomro last updated on 19/May/16

Region ABCDE = Region ABCDEO - ▴ AEO

In general, if first corner is A_{1} and xth corner

(upto wbich area is required and of course

0 < x < n for n - gon) is A_{x} then,

Region A_{1} A_{2} \dots A_{x} = Region A_{1} A_{2} \dots A_{x} O - ▴ A_{1} A_{x} O

Region A_{1} A_{2} \dots A_{n} O consists of x - 1 isosceles

triangles in general (only in hexagon these triangles

are equilateral) . If polygon is n - gon each of these

triangles has \frac{360}{n} degree angle at centre (O) of

polygon and △ A_{1} A_{x} O has (x - 1) (\frac{360}{n}) dgree angle

at O .

Let radius of the polygon is r

Each triangle in Region A_{1} A_{2} \dots A_{x} O has area :

▴ = \frac{1}{2} r^{2} \sin \frac{360}{n}

Region A_{1} A_{2} \dots A_{x} O consist of x - 1 such triangles

So,

Region A_{1} A_{2} \dots A_{x} O = (x - 1) \times \frac{1}{2} r^{2} \sin \frac{360}{n}

= \frac{1}{2} r^{2} (x - 1) \sin \frac{360}{n}

And now ▴ A_{1} A_{x} O = \frac{1}{2} r^{2} \sin \frac{360 (x - 1)}{n}

So finally

Region A_{1} A_{2} \dots A_{x} = Region A_{1} A_{2} \dots A_{x} O - ▴ A_{1} A_{x} O

= \frac{1}{2} r^{2} (x - 1) \sin \frac{360}{n} - \frac{1}{2} r^{2} \sin \frac{360 (x - 1)}{n}

= \frac{1}{2} r^{2} [(x - 1) \sin \frac{360}{n} - \sin \frac{360 (x - 1)}{n}]

Commented by FilupSmith last updated on 17/May/16

Very clever!!! thanks!