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Author: Subject: the area of star shape HELP PLS!!!
gorkem
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sad.gif posted on 5-2-2005 at 13:53
the area of star shape HELP PLS!!!


HELLO everybody
i would like to draw a 1cm² star shape.
will you please tell me the lenghts of the edges and the angels of the triangles. thank you very much for your help indeed.:(
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Mumbles
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[*] posted on 5-2-2005 at 16:54


1 cm2 total area of the star itself, or a star fitting inside of a 1 cm2 box?

If it's a box, make all lengths 1 cm.

Somebody correct me if this is wrong:

A congruent 5 pointed star happens to be associated with the golden ratio(1.618...). If you put a pentagon around a 5 pointed star the star takes up exactly 61.8% of the area. So by finding the area of the pentagon, and dividing by 1.618, or multiplying by .618(interesting how that works out isn't it?) you would find the are of the star inside.

All you have to do is find the dimensions of the pentagon that has an area of 1.618, then draw in the diagnals and make your star.

If you want to make it a little harder there is a formula for finding the area of a pentagram star which is a perfect 5 pointed star:

A=5s<sup>2</sup>tan( 3pi/10 )/4 - 5s<sup>2</sup>tan( pi/5 )/4

s is the side length of the pentagon, from which you can draw or calculate the length of the lines you need by using the law of cosines or sines or something.

Just FYI the angle between a regular pentagon and an inscribed triangle is 3pi/10 rad or 54 degrees. This makes the measure of each point of a star's angle equal 12 degrees.

[Edited on 2-6-2005 by Mumbles]
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JohnWW
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[*] posted on 5-2-2005 at 18:07


That can also be expressed as a derivative of the square root of 5. Tan(¶/5) = ¼(sqrt5 - 1)sqrt(2(5 - sqrt5)). Tan (3¶/10) = (1 + sqrt5)/sqrt(2(5 - sqrt5)) .

[Edited on 6-2-2005 by JohnWW]
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[*] posted on 5-2-2005 at 18:57


Since I don't want to figure out how to type phi's here, let p=(-1+sqrt(5))/2; this is about .618 as Mumbles said. When you draw a pentagram inscribe in a pentagon, you get a mix of two types of triangles, pointy ones and blunt one. Both have side ratio p. Examining the figure, the big pentagon is mage up of 1 small pentagon plus 5 each small pointy and small blunt triangles. Using the edge ratio of the triangles, the small pentagon has sides p^2 times those of the big one, so the area ratio is p^4. This is entirely within the pentagram. The remaining (1-p^4)th of the area is composed of small triangles, and the area ratio between a small pointy and the sum of a small pointy and a small blunt is p^2. Thus, the total area within the pentagram is p^2*(1-p^4)+p^4 times that of the whole pentagon. You really should be able to look up the area of a whole pentagon yourself, or use the formula JohnWW gave.

Mumbles: You didn't really mean to say 12 degrees, did you?
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[*] posted on 6-2-2005 at 11:02


No, I didn't mean to say 12. I knew that didn't look right. I messed up a number, and that caused a very small value for that. I was thinking the angle between the pentagram and the pentagon was 54 degrees, it is actually 36.

The angle of a perfect star corner is actually 36 degrees.
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