centroids in the normalised case. d e c i m a l p a r t ( 4.8284271... + 0.0 ) = d e c i m a l p a r t ( 4.8284271... ) = .8284271 {\displaystyle decimalpart(4...
the semicircle r 1 = ( 2 − 2 ) ⋅ s {\displaystyle r_{1}=(2-{\sqrt {2}})\cdot s\quad } See QH_dia 2) Radius of the additional circle r 2 = 2 − 2 2 ⋅ s...
} , see details under FS QV dia.png 2) Perimeter of the circle: P 2 = 2 π r 2 = 2 π ⋅ ( 2 − 1 ) ⋅ a {\displaystyle P_{2}=2\pi r_{2}=2\pi \cdot ({\sqrt...
⇔ r 0 r 4 = r 0 2 − 2 r 0 r 4 − 1 4 r 0 2 − r 0 r 4 {\displaystyle \Leftrightarrow r_{0}r_{4}=r_{0}^{2}-2r_{0}r_{4}-{\frac {1}{4}}r_{0}^{2}-r_{0}r_{4}\quad...
⇔ r 0 r 4 = r 0 2 − 2 r 0 r 4 − 1 4 r 0 2 − r 0 r 4 {\displaystyle \Leftrightarrow r_{0}r_{4}=r_{0}^{2}-2r_{0}r_{4}-{\frac {1}{4}}r_{0}^{2}-r_{0}r_{4}\quad...
|AF|^{2}+|FS_{2}|^{2}=(r_{1}+r_{2})^{2}} , applying equation (5) ⇔ | A F | 2 + r 2 2 = ( r 1 + r 2 ) 2 {\displaystyle \quad \Leftrightarrow |AF|^{2}+r...
{\displaystyle r_{1}} around point A {\displaystyle A} . From the developing of FS_CV we know that r 1 = 2 ⋅ r 0 {\displaystyle r_{1}={\sqrt {2}}\cdot r_{0}} ...
construction of FS_QH it is known that the line segment [ A K ] {\displaystyle [AK]} has the length of the radius r 1 {\displaystyle r_{1}} . This leads...
|AD|^{2}+|AB|^{2}=|BD|^{2}} ⇔ r 1 2 + r 1 2 = ( 2 ⋅ r 0 ) 2 {\displaystyle \quad \Leftrightarrow r_{1}^{2}+r_{1}^{2}=(2\cdot r_{0})^{2}} ⇔ 2 ⋅ r 1 2 = 4 ⋅ r 0 2 {\displaystyle...
2 ⋅ r 1 + π ⋅ r 1 = ( 2 + π ) ⋅ r 1 = ( 2 + π ) ⋅ ( 2 − 2 ) ⋅ a 0 ≈ 3.012 ⋅ a 0 {\displaystyle P_{1}=2\cdot r_{1}+\pi \cdot r_{1}=(2+\pi )\cdot r_{1}=(2+\pi...
_dia.png){kind=link}
