HELP!

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2 total messages Started by John Peacock Thu, 03 Jun 1999 00:00

#20

Author: John Peacock
Date: Thu, 03 Jun 1999 00:00

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DOES ANY ONE KNOW THE Nth TERM TO FIND THE AREA OF ANY REGULAR POLYGON

IF YOU DO PLEASE E-MAIL ME:

PEACOCKJOHN@tesco.net

thanks

Re: HELP!

#21

Author: Rob Morewood
Date: Sat, 05 Jun 1999 00:00

26 lines
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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