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R={\tfrac {h}{2}}+{\tfrac {c^{2}}{8h}}

^[1]

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\theta =2\arcsin {\tfrac {c}{2R}}

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c=2R\sin {\tfrac {\theta }{2}}=R{\sqrt {2(1-\cos \theta )}}

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h=R(1-\cos {\tfrac {\theta }{2}})=R\left(1-{\sqrt {\tfrac {1+\cos \theta }{2}}}\right)

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s={\theta }R

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a={\tfrac {R^{2}}{2}}\left(\theta -\sin \theta \right)

R, h ???????????? ???????????,

A=R^{2}\arccos \left(1-{\frac {h}{R}}\right)-\left(R-h\right){\sqrt {R^{2}-\left(R-h\right)^{2}}}

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↑ The fundamental relationship between R, c, and h derivable directly from the Pythagorean theorem among R, C/2 and r-h components of a right-angled triangle is: $R^{2}=({\tfrac {c}{2}})^{2}+(R-h)^{2}$ which may be solved for R, c, or h as required.

Weisstein, Eric W. "Circular segment". MathWorld.

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[1] The fundamental relationship between R, c, and h derivable directly from the Pythagorean theorem among R, C/2 and r-h components of a right-angled triangle is: $R^{2}=({\tfrac {c}{2}})^{2}+(R-h)^{2}$ which may be solved for R, c, or h as required.

[1]