What is a polygon with twelve sides
What is a regular polygon?
|The regular polygon has|
n equally long sides and
n interior angles of equal size. (The variable n stands for a natural number greater than 2.)
The heptagon is usually chosen as a substitute for n-corner for the drawings.
The regular polygon is also called a regular n-gon or, based on the English name, regular polygon. On this page it is simply called polygon.
The regular polygons triangle to Dodecagon are discussed individually elsewhere on my site.
You can access it from the main page.
Sizes of the polygon Top
|A regular n-gon is determined by the yellow triangle.|
The central angle is 360 ° / n.
The interior angle is (n-2) / n * 180 °.
The sum of all interior angles is (n-2) * 180 °.
Name of the diagonals
|......||It is useful to mark the diagonals of the polygon (as with the heptagon on the left) with di to call. One then accepts that di can also be one side and diagonals are called double.|
But this way the formulas below for the diagonals become simple.
|......||There are sizes|
page a, Radius r of the inscribed circle, radius R. of the perimeter, diagonals di, Area A. and scope U.
If side a is given, then applies
Derivation of the formulas
radius r of the inscribed circle and radius R. of the perimeter
|We have sin (180 / n) = (a / 2) / R. It follows R.= a / [2 * sin (180 / n)]|
The following applies: tan (180 / n) = (a / 2) / r. It follows r= a / [2 * tan (180 / n)].
|We have sin (180i / n) = (i.e.i/ 2) / R or di= 2R * sin (180 ° i / n).|
With R = a / [2 * sin (180 / n)] di= [a * sin (180i / n)] / sin (180 / n)
In the drawing, n = 7 and i = 3.
Area and perimeter
The area is n times the area of the determination triangle: A.= n * (a * r / 2) = na² / [4 * tan (180 / n)].
The scope is U= na.
Supplement: height h
If the polygon has an odd number of corner points, the "height" h lies on the axis of symmetry of the polygon.
|The following applies: tan (90 ° / n) = (a / 2) / h or H= a / [2tan (90 ° / n)].|
Number of diagonals Top
Position of the diagonals
|For example, there are different images for the ten and nine corners.|
If the number of corners is even, the diagonals run through the center point.
If the number of corners is odd, the center is empty and the diagonals form a reduced n-gon on the inside.
Number of all diagonals
If you specify the n points of the polygon, you can connect the first point with n-1 points.
Then you can connect the next point with n-2 points.
The next but one point is connected with n-3 points.
You continue this process until there is no point left to connect.
There are a total of (n-1) + (n-2) + (n-3) + ... + 3 + 2 + 1 connecting lines. That can be summed up with n (n-1) / 2.
Finally, you have to subtract the n lines between the corner points.
This leads to n (n-1) / 2-n = n (n-3) / 2 diagonals.
Result: The polygon has n (n-3) / 2 diagonals.
Number of diagonals of different lengths
|.....||The number of diagonals of different lengths is obtained if only one point is connected to all corner points.|
That gives n-1 routes.
Then you pull off the two sides. That makes n-3.
Finally halve the number and get (n-3) / 2.
This consideration only applies to polygons with an odd number of corners.
|......||If the number of corners of the polygon is even, then a diagonal is the axis of symmetry and occurs only once.|
Therefore the number is equal to (n-3 + 1) / 2 = (n-2) / 2 if n is even.
Every regular polygon is
> point symmetric (the center of symmetry is the center point.)
> n-fold rotationally symmetrical. (With a rotation of 360 ° / n, the polygon comes into congruence again.)
> mirror symmetry (there are n axes of symmetry.)
|......||If the number of corners is even, an axis of symmetry always runs through two corner points or through two side centers.|
|......||If the number of corners is odd, an axis of symmetry always runs through a corner and a side center.|
Division of the polygons Top
Division into triangles
It's about the problem of how many triangles the diagonals divide the polygon into.
|For the pentagon, for example, there are five decompositions.|
Mirror-image decompositions are viewed as different.
The number is generally indicated by the Catalan numbers:
Division into polygons
If you draw in all diagonals and also pay attention to their intersection points, the problem of dividing the regular polygon into different polygons arises.
|......||For the square there are 4, for the pentagon 11 and for the hexagon 24 polygons, as can be confirmed by counting.|
The first members of the sequence are 1,4,11,24,50,80,154,220,375,444 ...
|For odd n the formula is generally:||A (n) = (24 - 42n + 23n ^ 2 - 6n ^ 3 + n ^ 4) / 24|
All triangles in the polygon
|The square has 4 + 2 + 2 = 8 triangles.|
The first parts of the sequence are 1, 8, 35, 110, 287, 632, ... (Sloane's A006600).
Circle and polygon Top
|......||Is the circle a polygon?|
The answer is no. The polygon would have to have an infinite number of corners in order to become a circle. But infinity is not a number. - One side should have the length 0. Then it is no longer a page.
This consideration is fruitful for the circle calculation, because the formulas of the n-corner can be adopted for "n towards infinity" for the circle.
There are two quantities of interest for the circle with a given radius r, namely the area A 'and the perimeter U'. Since A 'and r² or U' and 2r are proportional and the proportionality factor is equal to Pi, the calculation of the circle amounts to determining Pi.
Circle between two polygons
|......||In a known method, one chooses a chordal polygon and a tangent polygon for a circle with the radius r and thus constrains the circle. For any n one determines the magnitudes un and Un of polygons and allows n to go beyond all limits.|
In the drawing, n = 6.
Determination triangles are sufficient for the calculation.
The following applies: sin (180 ° / n) = sn/ (2r) or sn= 2r * sin (180 ° / n)
The following applies: tan (180 ° / n) = Sn/ (2r) or S.n= 2r * tan (180 ° / n)
The scope applies un= n * sn= 2rn * sin (180 ° / n) and Un= n * Sn= 2rn * tan (180 ° / n).
Then the interval nesting applies
2rn * sin (180 ° / n) n * sin (180 ° / n) With the help of this chain of inequalities one can determine Pi as precisely as required.
Numerical example n = 100: 100sin1.8 °
Interval nesting is obtained for the areas
Numerical example n = 100: 50 * sin3.6 °
This way of determining pi is not used in school, also for a practical reason. Trigonometric functions are only dealt with after the circle calculation. Instead, pi is recorded using root terms.
You can also get to them via the polygons. One examines e.g. the 4-corner, 8-corner, 16-corner etc. one after the other.
The root terms are then calculated with a pocket calculator.
In a small volume from 1913 (1) I found the uncalculated terms for r = 1 from pre-computer times
Constructibility of the polygons Top
|......||The equilateral triangle can be drawn with a compass and ruler. This is called constructing.|
The triangle can also be used to construct the hexagon, dodecagon, 24-sided etc., because they are obtained one after the other by bisecting the angle.
The question arises whether one can construct all regular polygons.
The problem is solved, even if one does not specifically know all constructible polygons (*).
According to Gauss, we know that only the p-vertex (p is a prime number) can be constructed for which there are natural numbers k such that p = 2 ^ (2 ^ k) +1 are prime numbers.
These are the n-vertices with the vertices 3, 5, 17, 257 and 65537. No other numbers are known (*).
From these "Fermat primes" pi you can create new numbers through product formation: p1p2p3... ps.
Note that each of these numbers can be the starting number of a sequence of new numbers because of the halving above, you get with 2m p1p2p3... ps a general representation. Polygons with this number of corners can therefore be constructed.
Specifically, these are the numbers
3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48, 51, 60, 64, 68, 80, 85, 96, .. . (Sloane's A003401).
The n-vertices with the vertices 7, 9, 11, 13, 14, 18, 19, 21, 22, 23, 25, ... cannot be constructed.
There are approximate constructions for them.
A maximum property Top
|......||The theorem applies to regular polygons:|
Of all the n-vertices inscribed in a circle, the regular one has the largest area.
The proposition is obvious, but the proof is not very easy.
A mentally and textually carefully presented proof can be found in Rademacher / Toeplitz (4) on pages 9 to 14.
Polygons on the Internet Top
ORNAMENTICS IN FOUR QUARTERS
My geometric patterns from regular polygons
Regular polygon and slicing
Regular Polygon, Polygon, Constructible Polygon
1728 Software Systems
Regular polygon calculator
Dr Math (The Math Forum)
Eric W. Weisstein (MathWorld)
N. J. A. Sloane (On-Line Encyclopedia of Integer Sequences)
A000108 (Catalan Numbers)
Paul A. Gusmorino
Paul's Page of Pi
Golden Fields: A Case for the Heptagon
Steven E. Sommars - Tim Sommars
The Number of Triangles Formed by Intersecting Diagonals of a Regular Polygon
Regular polygon, Polygon, Constructible polygon
|Kenji Okamoto Polycircle ("We love LOGO")|
(1) Eugen Beutel: Squaring the Circle, Leipzig Berlin 1913/1942
(2) W. Lietzmann: Old and new from the district, Leipzig Berlin 1935
(3) Jean-Paul Delahaye: Pi die Story, Birkhäuser Verlag, Basel Boston Berlin 1999
(4) Rademacher-Toeplitz: Of Numbers and Figures, Springer, Berlin, Heidelberg, New York 1968 (reprint from 1930)
Feedback: Email address on my main page
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© 2005 Jürgen Köller
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