Pole wycinka koła o promieniu r i kącie 𝛼 obliczamy wzorem:

$P_w=\frac{\alpha }{{360}^{\circ }}\cdot \pi r^2$  

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$A=\frac{{165}^{\circ }}{{360}^{\circ }}\cdot \pi \cdot 6^2=\frac{11}{24}\cdot 36\pi =\frac{11}{2}\cdot 3\pi =\frac{33}{2}\pi$  

$B=\frac{{45}^{\circ }}{{360}^{\circ }}\cdot \pi \cdot 6^2=\frac{1}{8}\cdot 36\pi =\frac{1}{2}\cdot 9\pi =\frac{9}{2}\pi$  

$C=\frac{{360}^{\circ }-\left({165}^{\circ }+{45}^{\circ }\right)}{{360}^{\circ }}\cdot \pi \cdot {10}^2=\frac{{360}^{\circ }-{210}^{\circ }}{{360}^{\circ }}\cdot 100\pi =\frac{{150}^{\circ }}{{360}^{\circ }}\cdot 100\pi =\frac{5}{12}\cdot 100\pi =\frac{5}{6}\cdot 50\pi =\frac{250}{6}\pi$  

$D=\frac{{45}^{\circ }}{{360}^{\circ }}\cdot \pi \cdot {10}^2=\frac{1}{8}\cdot 100\pi =\frac{1}{2}\cdot 25\pi =\frac{25}{2}\pi$