Lunule (Ippocrate di Chio, V sec. a.C.)
L = π(r√2/2)²/2 - (πr²/4 - r²/2) = r²/2 = T
L = π(r√2/2)²/2 - (πr²/4 - r²/2) = r²/2 = T
https://latex.codecogs.com/svg.image?L=\pi\left(\frac{\sqrt{2}}{2}r\right)^2\frac{1}{2}-\left(\pi\frac{r^2}{4}-\frac{r^2}{2}\right)=\frac{r^2}{2}=T
L=\pi\left(\frac{\sqrt{2}}{2}r\right)^2\frac{1}{2}-\left(\pi\frac{r^2}{4}-\frac{r^2}{2}\right)=\frac{r^2}{2}=T