English: Largest circle in a equilateral triangle that contains a largest circle and a the further largest circle - Details: EC(3) dia.png
Deutsch: Größter Kreis in einem gleichseitigen Dreieck, das einen größten Kreis und dessen nächstgrößten Kreis enthält - Details: EC(3) dia.png
Дата
Источник
Собственная работа
Автор
Hans G. Oberlack
The equilateral triangle as base element.
Inscribed is the largest circle.
Inscribed is the next largest circle.
Inscribed is the next largest circle.
General case
Segments in the general case
0) The side length of the equilateral base triangle is:
1) The radius of the circle is: , see calculation 3
2) The radius of the circle is: , see calculation 4
3) The radius of the circle is: , for symmetry reasons
Perimeters in the general case
0) Perimeter of equilateral base triangle:
1) Perimeter of inscribed circle:
2) Perimeter of inscribed circle:
3) Perimeter of inscribed circle:
Areas in the general case
0) Area of the equilateral base triangle: , see calculation (2)
1) Area of the inscribed circle:
2) Area of the inscribed circle:
2) Area of the inscribed circle:
Covered surface of base shape:
Centroids in the general case
1) Centroids as graphically displayed
Centroid positions are measured from the centroid point of the base shape
0) Centroid position of the base square:
1) Centroid position of the inscribed circle:
2) Centroid of the additional circle:
3) Centroid of the additional circle: , see Calculation (6)
2) Orientated centroids
The centroid positions of the following shapes will be expressed orientated so that the first shape n with will be of type with . This means that the graphical representation will not correspond to the mathematical expression.
0) Orientated centroid position of the base circle:
1) Orientated centroid position of the inscribed circle:
2) Orientated centroid of the additional circle:
3) Orientated centroid of the additional circle: , see Calculation (6)
Normalised case
In the normalised case the area of the base shape is set to 1.
So
Segments in the normalised case
0) Side length of the triangle
1) The radius of the circle is: ,
2) The radius of the additional circle is:
2) The radius of the additional circle is:
Perimeters in the normalised case
0) Perimeter of base triangle:
1) Perimeter of inscribed circle:
2) Perimeter of inscribed circle:
3) Perimeter of inscribed circle:
Areas in the normalised case
0) Area of the base triangle is by definition
1) Area of the inscribed circle:
2) Area of the inscribed circle:
3) Area of the inscribed circle:
Centroids in the normalised case
1) Centroids as graphically displayed
Centroid positions are measured from the centroid point of the base shape
0) Centroid position of the base square:
1) Centroid position of the inscribed circle:
2) Centroid of the additional circle:
3) Centroid of the second additional circle:
2) Orientated centroids
The centroid positions of the following shapes will be expressed orientated so that the first shape n with will be of type with . This means that the graphical representation will not correspond to the mathematical expression.
0) Orientated centroid position of the base circle:
1) Orientated centroid position of the inscribed circle:
2) Orientated centroid of the additional circle:
3) Centroid of the second additional circle:
Calculations
Known elements
(0) Given is the side length of the equilateral triangle: (1) (2) (3) (4) (5) (6)
Calculation 1
The height is calculated: ,applying the Pythagorean theorem on the rectangular triangle , applying equation (2) , applying equation (1) , rearranging , rearranging , rearranging
Calculation 2
, applying equation (2) , applying result of calculation (2)
To calculate we use the two right triangles and , applying the Pythagorean theoreme on , applying equation (2) , applying equation (1) , rearranging , multiplying , rearranging , breaking into parts , applying equation (3) , applying equation (6) , applying equation (5) , rearranging , see calculation (5) , rearranging , extracting the roots , rearranging , applying calculation (3) , expanding , rearranging , rearranging , rearranging
Calculation 5
In order to find the size of the right triangle is used: , applying equation (5) , applying equation (4) , applying the Sinus function , rearranging , rearranging
Calculation 6
In order to find the position of centroid the position of centroid is rotated by , since the triangle is an equilateral triangle. The rotation by corresponds to a multiplication with so this gives:
a) In the general case as displayed: , since
b) In the orientated general case: , since
Лицензирование
Я, владелец авторских прав на это произведение, добровольно публикую его на условиях следующей лицензии:
делиться произведением – копировать, распространять и передавать данное произведение
создавать производные – переделывать данное произведение
При соблюдении следующих условий:
атрибуция – Вы должны указать авторство, предоставить ссылку на лицензию и указать, внёс ли автор какие-либо изменения. Это можно сделать любым разумным способом, но не создавая впечатление, что лицензиат поддерживает вас или использование вами данного произведения.
https://creativecommons.org/licenses/by/4.0CC BY 4.0 Creative Commons Attribution 4.0 truetrue
Краткие подписи
Добавьте однострочное описание того, что собой представляет этот файл
Largest circle in a equilateral triangle that contains a largest circle and a the further largest circle
Größter Kreis in einem gleichseitigen Dreieck, das einen größten Kreis und dessen nächstgrößten Kreis enthält
Файл содержит дополнительные данные, обычно добавляемые цифровыми камерами или сканерами. Если файл после создания редактировался, то некоторые параметры могут не соответствовать текущему изображению.