Benutzer:Hans G. Oberlack/QC 1.1415927

Shows the largest semicircle within a square.

General case

Base is the square $[ABCD]$ of side length s.

The radius r of the inscribed circle has the length ${\frac {s}{2}}$

Segments in the general case

0) The side length of the base square $=s$
1) Radius of the circle $r={\frac {s}{2}}$

Perimeters in the general case

0) Perimeter of base square$=4\cdot s$
1) Perimeter of the circle $=2\cdot \pi \cdot r=\pi \cdot s$

Areas in the general case

0) Area of the base square $=s^{2}$
1) Area of the inscribed circle $=\pi \cdot r^{2}=\pi \cdot ({\frac {s}{2}})^{2}=\pi \cdot {\frac {s^{2}}{4}}={\frac {\pi }{4}}\cdot s^{2}$

Centroids in the general case

Centroid positions are measured from the lower left point of the square.
0) Centroid positions of the base square: $x_{c}={\frac {s}{2}}\quad y_{c}={\frac {s}{2}}$
1) Centroid positions of the inscribed circle: $x_{c}={\frac {s}{2}}\quad y_{c}={\frac {s}{2}}$

Normalised case

In the normalised case the area of the base is set to 1.
$||ABCD||=1\Rightarrow s^{2}=1\Rightarrow s=1$

Segments in the normalised case

0) Segment of the base square $s=1$
1) Segment of the inscribed circle $r={\frac {1}{2}}=0.5$

Perimeters in the normalised case

0) Perimeter of base square$=4$
1) Perimeter of the inscribed circle $=2\cdot \pi \cdot r=2\cdot \pi \cdot {\frac {1}{2}}=\pi =3.1415926...$

Areas in the normalised case

0) Area of the base square $=1$
1) Area of the inscribed circle $=\pi \cdot r^{2}=\pi \cdot ({\frac {1}{2}})^{2}={\frac {\pi }{4}}=0.785398...$

Centroids in the normalised case

Centroid positions are measured from the lower left point of the square.
0) Centroid positions of the base square: $x_{c}={\frac {1}{2}}\quad y_{c}={\frac {1}{2}}$
1) Centroid positions of the inscribed circle:$x_{c}={\frac {1}{2}}\quad y_{c}={\frac {1}{2}}$