Area with Lunar Moons (2)

If the legs of the triangle are $a$ and $b$ , then the area of the lunes is the two semicircles, $\frac{\pi}2a^2+\frac{\pi}2b^2$ , plus the triangle $\frac{ab}2$ , minus the white semicircle $\frac{\pi}2c^2$

But $\frac{\pi}2a^2+\frac{\pi}2b^2+\frac{ab}2-\frac{\pi}2c^2=\frac{\pi}2(a^2+b^2-c^2)+\frac{ab}2=\frac{\pi}2\times0+\frac{ab}2=\frac{ab}2$