P_w- pole szukanego wycinka koła

l_w-długość odpowiadającego mu  łuku 

$a)$  

$\pi r^2=\pi \cdot {\left(\frac{10}{3}\right)}^2=\frac{100}{9}\pi$  

$P_w=\frac{{270}^{\circ }}{{360}^{\circ }}\cdot \left(\frac{100}{9}\right)\pi =\frac{300}{36}\pi =\underline{\frac{25}{3}\pi }$ $P_w=\frac{{270}^{\circ }}{{360}^{\circ }}\cdot \left(\frac{100}{9}\right)\pi =\frac{300}{36}\pi =\underline{\frac{25}{3}\pi }$   

$l_w=\frac{3}{4}\cdot 2\pi r=\frac{3}{4}\cdot 2\pi \frac{10}{3}=\underline{5\pi }$       

$b)$  

$P_w=\frac{{135}^{\circ }}{{360}^{\circ }}\cdot \pi r^2=\frac{135}{360}\cdot \pi \cdot 25=\frac{675}{72}\pi =\underline{\frac{75}{8}\pi }$    

$l_w=\frac{135}{360}\cdot 2\pi r=\frac{135}{360}\cdot 10\pi =\frac{135}{36}\pi =\underline{\frac{15}{4}\pi }$  

$c)$  

$P_w=\frac{39}{360}\cdot \pi r^2=\frac{39}{360}\cdot {\left(\frac{9}{2}\right)}^2\pi =\frac{39}{360}\cdot \frac{81}{4}\cdot \pi =\underline{\frac{351}{160}\pi }$          

$l_w=\frac{39}{360}\cdot 2\pi \cdot r=\frac{39}{180}\cdot \frac{9}{2}\cdot \pi =\underline{\frac{39}{40}\pi }$