$\pi r^2\cdot \frac{1}{2}=8\pi$   

$\pi r^2=16\pi$  

$r^2=16$  

$r=4$  

$\text{Suma kątoˊw wewnętrznych dowolnego czworokąta jest roˊwna}{360}^{\circ }.$  

${90}^{\circ }+{90}^{\circ }+\alpha +\gamma ={360}^{\circ }$  

$\alpha +\gamma ={180}^{\circ }$  

 

$\delta ={180}^{\circ }-\alpha$  

${90}^{\circ }+{90}^{\circ }+\delta +\beta ={360}^{\circ }$  

$\delta +\beta ={180}^{\circ }$  

${180}^{\circ }-\alpha +\beta ={180}^{\circ }$  

$\alpha =\beta$  

$\delta =\gamma$  

$\text{Czyli czworokąty AOED i BCEO są podobne.}$  

$\frac{P_{AOED}}{P_{BCEO}}=k^2=4$  

$k=2$  

$\frac{\left|DE\right|}{\left|EO\right|}=2$  

$\left|DE\right|=2\cdot 4=8$  

$\left|AD\right|=\left|DE\right|=8$  

$\left|CB\right|=\frac{1}{2}\left|AO\right|=2$  

 

$\underline{P_{ABCD}=\frac{8+2}{2}\cdot 2\cdot 4=40}$  

 

${\text{Szukane pole wynosi 40 dm}}^2.$