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.


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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

	$Region ABCDE = Region ABCDEO - ▴ AEO$ $In general, if first corner is A_{1} and xth corner$ $(upto wbich area is required and of course$ $0 < x < n for n - gon) is A_{x} then,$ $Region A_{1} A_{2} . . . A_{x} = Region A_{1} A_{2} . . . A_{x} O - ▴ A_{1} A_{x} O$ $Region A_{1} A_{2} . . . A_{n} O consists of x - 1 isosceles$ $triangles in general (only in hexagon these triangles$ $are equilateral) . If polygon is n - gon each of these$ $triangles has \frac{360}{n} degree angle at centre (O) of$ $polygon and △ A_{1} A_{x} O has (x - 1) (\frac{360}{n}) dgree angle$ $at O .$ $Let radius of the polygon is r$ $Each triangle in Region A_{1} A_{2} . . . A_{x} O has area :$ $▴ = \frac{1}{2} r^{2} \sin \frac{360}{n}$ $Region A_{1} A_{2} . . . A_{x} O consist of x - 1 such triangles$ $So,$ $Region A_{1} A_{2} . . . A_{x} O = (x - 1) \times \frac{1}{2} r^{2} \sin \frac{360}{n}$ $= \frac{1}{2} r^{2} (x - 1) \sin \frac{360}{n}$ $And now ▴ A_{1} A_{x} O = \frac{1}{2} r^{2} \sin \frac{360 (x - 1)}{n}$ $So finally$ $Region A_{1} A_{2} . . . A_{x} = Region A_{1} A_{2} . . . A_{x} O - ▴ A_{1} A_{x} O$ $= \frac{1}{2} r^{2} (x - 1) \sin \frac{360}{n} - \frac{1}{2} r^{2} \sin \frac{360 (x - 1)}{n}$ $= \frac{1}{2} r^{2} [(x - 1) \sin \frac{360}{n} - \sin \frac{360 (x - 1)}{n}]$
	Commented by FilupSmith last updated on 17/May/16

	$Very clever!!! thanks!$
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