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Article #21Re: HELP!
From: Rob Morewood
Date: Sat, 05 Jun 1999 00:00
Date: Sat, 05 Jun 1999 00:00
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1623 bytes
John Peacock (Peacockjohn@tesco.net) wrote:
: DOES ANY ONE KNOW THE Nth TERM TO FIND THE AREA OF ANY REGULAR POLYGON
To find the area of a regular N-gon:
Divide the polygon into 2N congruent right triangles by cutting from the
center to each vertex and to the midpoint of each side. (Draw a picture.)
The central angle of each triangle is 360 degrees divided by 2N (or 180/N).
The opposite side is half the side of the original polygon (S/2).
A little trig finds the adjacent side: (S/2)/Tan(180/N).
So the area of each triangle is half the base time the height:
(S/2)^2 / Tan(180/N)
The total area of the 2N triangles becomes: NS^2/(2Tan(180/N))
This analysis assumes that you know the number of sides N and the length
of each side S. If you had instead, for example, the apothesis, R,
(distance from center to the midpoint of a side) the formula becomes:
NR^2*Tan(180/N)
For large N, Tan(180/N) just slighly over PI/N, so the area is just over:
NR^2*PI/N = PI*R^2
|\/| || Burnaby South Secondary School
| |orewood@planeteer.com || Beautiful British Columbia
Mathematics & Computer Science || (Canada)
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X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: f7972,295774b9818e8688 X-Google-Attributes: gidf7972,public From: see.sig@for.real.address (Rob Morewood) Subject: Re: HELP! Date: 1999/06/05 Message-ID: <19990605.73111C8.2F73@van4p17.planeteer.com> X-Deja-AN: 485978727 References: <7j5toc$abd$2@epos.tesco.net> Organization: UUNET Canada News Transport Newsgroups: can.schoolnet.math.sr