Pole wycinka koła:

$P_1=\frac{90^{\circ }}{360^{\circ }}\cdot \pi \cdot {\left(\frac{\sqrt{2}}{2}\right)}^2=\frac{1}{4}\cdot \pi \cdot \frac{2}{4}=\frac{\pi }{8}\left[{\text{cm}}^2\right]$

Pole trójkąta $AOB$:

$P_2=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2}}{2}=\frac{2}{8}=\frac{1}{4}\left[{\text{cm}}^2\right]$

Pole figury:

$P=P_1-P_2=\left(\frac{\pi }{8}-\frac{1}{4}\right)\left[{\text{cm}}^2\right]$

Odp:. Pole zaznaczonej figury wynosi $\left(\frac{\pi }{8}-\frac{1}{4}\right){\text{cm}}^2$.