GeoGebra tutorial: Regular polygon with given length and center at a given point

Example: Regular triangle with a side length of 12 centered at the origin

Step: Compute the circumradius. In an equilateral triangle, s=R312=R3R=43s=R\sqrt{3} \rightarrow 12 = R\sqrt{3} \rightarrow R = 4\sqrt{3}

Step: Plot vertex A. A sits on the positive y-axis at distance R=43R=4\sqrt{3}.

A=(0, 4*sqrt(3))

Step: Plot vertex B and C. They extend symmetrically to the left and right by half the side length: 122=6\frac{12}{2} = 6. The distance from the center to the flat bottom side is the apothem, which for an equilateral triangle is exactly half of the circumradius: ap3=R2=432=23ap_{3} = \frac{R}{2} = \frac{4\sqrt{3}}{2} = 2\sqrt{3}. Since it is below the x-axis, it is negative.

B = (6, -2\sqrt{3})
C = (-6, -2\sqrt{3})

Step: Draw the polygon

triangle = Polygon(A, B, C)

Example: Regular hexagon with a side length of 12 centered at the origin

Step: In a regular hexagon, the circumradius equals the side length, so the diameter is two times the circumradius, so vertices on the horizontal axis (A and D) sit exactly at D(-12, 0) and A(12, 0)

A = (12, 0)
D = (-12, 0)

Step: Plot remaining vertices B, C, E and F. The remaining vertices are shifted horizontally by half the radius (R2=122=6\frac{R}{2}=\frac{12}{2}=6) and vertically by the apothem height ap6=R23=1223=63ap_{6} = \frac{R}{2}\sqrt{3}=\frac{12}{2} \sqrt{3} = 6\sqrt{3}

  • \(B = (6, 6\sqrt{3})\) (Top Right)
  • \(C = (-6, 6\sqrt{3})\) (Top Left)
  • \(E = (-6, -6\sqrt{3})\) (Bottom Left)
  • \(F = (6, -6\sqrt{3})\) (Bottom Right)
B = (6, 6*sqrt(3))
C = (-6, 6*sqrt(3))
E = (-6, -6*sqrt(3))
F = (6, -6*sqrt(3))

Step: Draw the polygon.

hexagon = Polygon(A, B, C, D, E, F)

Example: Regular dodecagon with a side length of 12 centered at the origin

Step: Compute the circumradius.

You can use any of the following expressions:

  • l12=R23R=1223l_{12}=R\sqrt{2-\sqrt{3}} \rightarrow R=\frac{12}{\sqrt{2-\sqrt{3}}}
  • l12=2Rsin(15)R=122sin(15)R=66+62l_{12}=2R\cdot\sin(15) \rightarrow R = \frac{12}{2\cdot sin(15)}\rightarrow R=6\sqrt{6} + 6\sqrt{2}
R = 6*sqrt(6) + 6*sqrt(2)

Step: Draw vertices at the x-axis and y-axis.

  • Vertex A ($\theta = 0^\circ$): Lands on the positive $x$-axis $\rightarrow (R, 0)$
  • Vertex D ($\theta = 90^\circ$): Lands on the positive $y$-axis $\rightarrow (0, R)$
  • Vertex G ($\theta = 180^\circ$): Lands on the negative $x$-axis $\rightarrow (-R, 0)$
  • Vertex J ($\theta = 270^\circ$): Lands on the negative $y$-axis $\rightarrow (0, -R)$
A = (R, 0)
D = (0, R)
G = (-R, 0)
J = (0, -R)

Step: Draw remaining vertices

B = (R*cos(30º), R*sin(30º))
C = (R*cos(60º), R*sin(60º))
E = (R*cos(120º), R*sin(120º))
F = (R*cos(150º), R*sin(150º))
H = (R*cos(210º), R*sin(210º))
I = (R*cos(240º), R*sin(240º))
K = (R*cos(300º), R*sin(300º))
L = (R*cos(330º), R*sin(330º))

Step: Draw polygon.

dodecagon = Polygon(A, B, C, D, E, F, G, H, I, J, K, L)
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