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## can one determine the surf/circumf with a geometrical figure

Exists in mathematics a geometric figure, with straight lines (rectangle, parallelogram, or so), allowing you to determine the circumference and area of a circle by drawing / calculating / approximating?
Without pi or formulas. So figure with no curved lines (and no polygons).
for instance using sin, cos, tan.

In other words: with pen, paper and a protractor drawing a figure to determine the circumference and surface(area) of a circle.

PS.
from other forums I heard it isn't possible...
BUT IT IS!!

I've found one model myself! It's a kite! With angle 144.686

For more detail, drawings and calculations view my blogsite:
http://quarks-divided.over-blog.fr
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 Recognitions: Homework Help Welcome to PF; http://a10.idata.over-blog.com/500x3...e-Circle-1.JPG That figure shows that when the angle at C is a particular value, then the area of the kite is the area of the circle. The calculation of the angle needs pi (and formulas).
 Mentor Blog Entries: 8 Right. Of course it is possible. But the idea is to construct that figure with compass and straightedge from the original circle. THAT is what is impossible.

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## can one determine the surf/circumf with a geometrical figure

Simon thanks for welcoming me!

I have never seen a circle being kited!

I added some new drawings, (logically):
"rectangling the circle"
and
"boxing the circle".

The big question is: why this angle of 17.65678715141......???????????????
 Recognitions: Homework Help The big question is: why this angle of 17.65678715141......??????????????? because the angles inside the triangle ABC must sum to 180 degrees. and those are the angles that make the area of the kite the same as the area of the circle. You get this kind of construction where you want to know (say) what the lowest point of reflection can be to reflect a signal between two places on the surface of the Earth.
 Recognitions: Gold Member because the angles inside the triangle ABC must sum to 180 degrees I understand yes. But why 17.686 and not 15.3 or 26.88 What's the relation of 17.6...to a circle or pi or so?
 Recognitions: Gold Member Science Advisor Staff Emeritus The angle is NOT necessarily 17.686. There are an infinite number of kites, of different dimensions and different angles that will work. You happen to have found one.
 Recognitions: Gold Member Thank you HallsofIvy for answering. But I can't believe you, sorry. I think there's just 1 kite: The kite which has 2 arms equal r and 2 arms equal πr and an area equal the area of a circle. Other angles can't give that I think.(I tried several) If you can give me an example of another kite with those 3 arguments, please! If not, why 17.686...?

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 Quote by Benedictijn because the angles inside the triangle ABC must sum to 180 degrees I understand yes. But why 17.686 and not 15.3 or 26.88
Because none of those angles will make the sum come to 180 degrees.

The key to the figure is the angle at C.
Here is how the kite is determined:
A and B are on the circle, C is in the center, Point D is the intersection between the tangents to the circle at A and B.

Angles: A+B+C+D=360, A=B=90, so D=180-C
... so setting angle C will determine the entire geometry of the kite. [1]

You want to find the kite whose area is also the area of the circle ... namely ##\frac{1}{2}|CD||AB|=\pi r^2##. To find the angle C that does this, you need to express |AB| and |CD| in terms of sines and cosines of C.

##|AB| = 2r\sin(C/2)## .. that was easy! The other one takes a bit more work but you should be able to do it OK.

But you needn't bother - ##C/2## is the angle ##\theta## in a right-angle triangle whose adjacent side is length ##r## and opposite side length ##\pi r##
The area of the kite is two of these.

It is easy to find shapes whose significant feature is also the area of a circle
... eg: the area of an oblong width pi and length r2 is also the area of a circle radius r;
... the volume of a box whose base is a square with sides length r, and of height pi, is the area of a circle radius r too.
... as above, the area of a box length ##\pi r## and width ##r##: cut that box corner-to-corner to make two triangles - flip one and put the hypotenuses back together and you have the above kite.

---------------------

[1] per your question ... if a = ∠CAB = ∠CBA, then 2a= 180-C or a=90-(C/2)
i.e. the angle is 17.686degrees because of what C has to be (by above).
 Recognitions: Gold Member Hello Simon, thanks for the extended answer. I think you should look at my blogsite with the drawings and calculations to see what I ment. We're talking 2 directions. It's not a box with edges pi*r and r but those with cos and cos/tan edges. I think my representation of the kite is the only one. Or not? Thanks see: quarks-divided.over-blog.fr and to the "kyte-ing the circle"page / "rectangling the circle"page
 Recognitions: Homework Help I did see your blog site... from the first post right? Your kite shape - cut it along the CD line to get two triangles, flip one of them over, put them back together. See? Each triangle of the kite is just a rectangle cut corner-to-corner. Using the figure on http://quarks-divided.over-blog.fr/ The area of the kite is the area of the two triangles CAD and CBD added together, which is twice the area of one of them since they are mirror images. The triangles have base length ##|AD|=|BD|=\pi r## (half the circumference) and height ##|CA|=|CB|=r## for area ##\frac{1}{2}\pi r^2## and twice that is??? The kite you describe is not the only one that can be drawn by those rules, but it is the only one whose area is the same as the area of the circle. You cannot construct it using a straight-edge and a protractor without also knowing the angle at C ... which you have to compute from knowing the value of pi and the tangent function. It has to be the one that makes the sides ##|AD|=\pi|CA|## and that is a tangent ratio.
 Recognitions: Gold Member to Simon: but it is the only one whose area is the same as the area of the circle !!!! That has been my statement all allong! The question is: why is that so and why does it give that kite with angle 17.656...(or angle C 144.686...) What's the relationship between that (those) angle(s) and a circle (or pi, or...). i.o.w. where does this number 17.65678715141....come from?? http://quarks-divided.over-blog.fr/p..._-8348481.html http://quarks-divided.over-blog.fr/p...e-8354341.html

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Homework Help
 The question is: why is that so and why does it give that kite with angle 17.656...(or angle C 144.686...)
I'm sorry I keep telling you - because the geometry dictates it. The angles have those (approximate) values for the same reason that 3-4-5 triangle has angles (approximately) 36.86° and 53.13°; it is for the same reason your legs are exactly long enough for your feet to touch the ground!

One... last... time... and since you seem to be fixated on the 17thingy number I'll do it that way this time:

17.65678715141...... is the half-angle at D, so ##D/2=17.65678715141......## (Using my convention of labeling interior angles with the point name.)

So it is the angle of a right-triangle whose adjacent side is length ##|DA| = \pi r## and whose opposite side is length ##|AC|=r## .... so ##\tan(D/2) = 1/\pi## because of the definition of the tangent function.

The sides have to have those lengths because one of them is defined to be the radius (so that's fixed) and the other side has to be long enough that the product of the sides is ##\pi r^2##.

Perhaps this will help:
If you hadn't set the angles at A and B to be 90°, then all these numbers would be different.
eg. if you had set the angle at C to 60°, then ##|AB|=r##, and ##|DC|=2\pi r##
... and the half-angle at D would be

##\tan(\frac{D}{2})=1/(4\pi-\sqrt{3})\Rightarrow D/2 \approx 5.2734^\circ##.

This method is more useful in terms of construction because you do the following:
1. draw a circle with radius r - mark the center
2. construct an equilateral triangle sides length r with one corner on the center - mark the other two corners
3. bisect the angle at the center - draw a line radially along the bisector
4. cut out a circle radius r to make a wheel
5. roll the wheel along the radial line until it turns exactly once and mark that point
6. join each of the marked points to form a kite-shape.
... the area of the kite is the area of the circle.
 Recognitions: Gold Member Well, to make a long story short: the answer to why 17.656... is because atan (1/pi) = 17.65678715141288.. (thanks to Simon.)