Summary
The equilateral triangle as base element.
General case
Segments in the general case
0) The side length of the equilateral base triangle is:
a
0
{\displaystyle a_{0}}
Perimeters in the general case
0) Perimeter of equilateral base triangle:
P
0
=
3
⋅
a
0
{\displaystyle P_{0}=3\cdot a_{0}}
Areas in the general case
0) Area of the equilateral base triangle:
A
0
=
3
4
⋅
a
0
2
≈
0.433
⋅
a
0
2
{\displaystyle A_{0}={\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}\quad \approx 0.433\cdot a_{0}^{2}\quad }
, see calculation (2)
Centroids in the general case
By definition the centroid point of a base shape is
S
0
=
0
+
0
i
{\displaystyle S_{0}=0+0i}
The positon of point D relative to
S
0
{\displaystyle S_{0}}
is:
D
=
0
−
1
3
⋅
|
C
D
|
⋅
i
=
0
−
1
3
⋅
3
2
⋅
a
0
⋅
i
=
(
0
−
1
2
3
⋅
i
)
⋅
a
0
{\displaystyle D=0-{\frac {1}{3}}\cdot |CD|\cdot i=0-{\frac {1}{3}}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{0}\cdot i=\left(0-{\frac {1}{2{\sqrt {3}}}}\cdot i\right)\cdot a_{0}}
Normalised case
In the normalised case the area of the base shape is set to 1.
So
A
0
=
3
4
⋅
a
0
2
=
1
⇒
a
0
2
=
4
3
⇒
a
0
=
2
3
≈
1.52
{\displaystyle A_{0}={\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}=1\quad \Rightarrow a_{0}^{2}={\frac {4}{\sqrt {3}}}\quad \Rightarrow a_{0}={\frac {2}{\sqrt {\sqrt {3}}}}\quad \approx 1.52}
Segments in the normalised case
0) Side length of the triangle
a
0
=
2
3
≈
1.52
{\displaystyle a_{0}={\frac {2}{\sqrt {\sqrt {3}}}}\quad \approx 1.52}
Perimeter in the normalised case
0) Perimeter of base triangle:
P
0
=
3
⋅
2
3
=
12
3
≈
4.5590141
{\displaystyle P_{0}=3\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\sqrt {12{\sqrt {3}}}}\quad \approx 4.5590141}
Area in the normalised case
0) Area of the base triangle is by definition
A
0
=
3
4
⋅
(
2
3
)
2
=
1
{\displaystyle A_{0}={\frac {\sqrt {3}}{4}}\cdot \left({\frac {2}{\sqrt {\sqrt {3}}}}\right)^{2}=1}
Centroids in the normalised case
By definition the centroid point of a base shape is
S
0
=
0
+
0
i
{\displaystyle S_{0}=0+0i}
The positon of point D relative to
S
0
{\displaystyle S_{0}}
is:
D
=
(
0
−
1
2
3
⋅
i
)
⋅
2
3
=
0
−
1
3
3
⋅
i
≈
0
−
0.439
i
{\displaystyle D=\left(0-{\frac {1}{2{\sqrt {3}}}}\cdot i\right)\cdot {\frac {2}{\sqrt {\sqrt {3}}}}=0-{\sqrt {\frac {1}{3{\sqrt {3}}}}}\cdot i\quad \approx 0-0.439i}
Calculations
Known elements
(0) Given is the side length of the equilateral triangle:
a
0
{\displaystyle a_{0}}
(1)
|
A
B
|
=
|
B
C
|
=
|
A
C
|
=
a
0
{\displaystyle \quad |AB|=|BC|=|AC|=a_{0}}
(2)
|
A
D
|
=
|
B
D
|
=
1
2
⋅
a
0
{\displaystyle \quad |AD|=|BD|={\frac {1}{2}}\cdot a_{0}}
Calculation 1
The height
|
C
D
|
{\displaystyle |CD|}
is calculated:
|
A
D
|
2
+
|
C
D
|
2
=
|
A
C
|
2
{\displaystyle \quad |AD|^{2}+|CD|^{2}=|AC|^{2}\quad }
,applying the Pythagorean theorem on the rectangular triangle
△
A
B
C
{\displaystyle \triangle ABC}
⇔
(
1
2
⋅
a
0
)
2
+
|
C
D
|
2
=
|
A
C
|
2
{\displaystyle \quad \Leftrightarrow \left({\frac {1}{2}}\cdot a_{0}\right)^{2}+|CD|^{2}=|AC|^{2}\quad }
, applying equation (2)
⇔
(
1
2
⋅
a
0
)
2
+
|
C
D
|
2
=
a
0
2
{\displaystyle \quad \Leftrightarrow \left({\frac {1}{2}}\cdot a_{0}\right)^{2}+|CD|^{2}=a_{0}^{2}\quad }
, applying equation (1)
⇔
|
C
D
|
2
=
a
0
2
−
1
4
⋅
a
0
2
{\displaystyle \quad \Leftrightarrow |CD|^{2}=a_{0}^{2}-{\frac {1}{4}}\cdot a_{0}^{2}\quad }
, rearranging
⇔
|
C
D
|
2
=
3
4
⋅
a
0
2
{\displaystyle \quad \Leftrightarrow |CD|^{2}={\frac {3}{4}}\cdot a_{0}^{2}\quad }
, rearranging
⇔
|
C
D
|
=
3
2
⋅
a
0
{\displaystyle \quad \Leftrightarrow |CD|={\frac {\sqrt {3}}{2}}\cdot a_{0}\quad }
, rearranging
Calculation 2
A
0
=
|
A
D
|
⋅
|
C
D
|
{\displaystyle \quad A_{0}=|AD|\cdot |CD|}
⇔
A
0
=
1
2
⋅
a
0
⋅
|
C
D
|
{\displaystyle \quad \Leftrightarrow A_{0}={\frac {1}{2}}\cdot a_{0}\cdot |CD|\quad }
, applying equation (2)
⇔
A
0
=
1
2
⋅
a
0
⋅
3
2
⋅
a
0
{\displaystyle \quad \Leftrightarrow A_{0}={\frac {1}{2}}\cdot a_{0}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{0}\quad }
, applying result of calculation (2)
⇔
A
0
=
3
4
⋅
a
0
2
≈
0.433
⋅
a
0
2
{\displaystyle \quad \Leftrightarrow A_{0}={\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}\quad \approx 0.433\cdot a_{0}^{2}\quad }
Licensing
I, the copyright holder of this work, hereby publish it under the following license:
You are free:
to share – to copy, distribute and transmit the work
to remix – to adapt the work
Under the following conditions:
attribution – You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.
share alike – If you remix, transform, or build upon the material, you must distribute your contributions under the same or compatible license as the original. https://creativecommons.org/licenses/by-sa/4.0 CC BY-SA 4.0 Creative Commons Attribution-Share Alike 4.0 true true English Equilateral triangle
German Gleichseitiges Dreieck