$\text{Wzoˊr na obwoˊd koła jest dany wzorem:}$  

$Obw=2\cdot \pi \cdot r$

 

 

$\text{Wzoˊr na pole koła jest dany wzorem:}$  

$P=\pi \cdot r^2$

 

 

$\text{Sˊrednica jest roˊwna podwojonemu promieniowi}$  

$d=2r$  

$a)d=3mm$

$\text{a więc}r=1,5mm$ $\text{a więc}r=1,5mm$  

$P=\pi \cdot {\left(1,5\right)}^2=\pi \cdot 2,25=2,25\pi \left[mm\right]$  

$Obw=2\cdot \pi \cdot 1,5=2\cdot 1,5\cdot \pi =3\pi \left[mm\right]$  

 

 

$b)d=\frac{2}{3}km$  

$\text{a więc}r=\frac{1}{3}km$

$P=\pi \cdot {\left(\frac{1}{3}\right)}^2=\pi \cdot \frac{1}{9}=\frac{1}{9}\pi \left[km\right]$

$Obw=2\cdot \pi \cdot \frac{1}{3}=2\cdot \frac{1}{3}\cdot \pi =\frac{2}{3}\pi \left[km\right]$

 

 

$c)d=\frac{\sqrt{3}}{2}mm$  

$\text{a więc}r=\frac{\sqrt{3}}{4}mm$  

$P=\pi \cdot {\left(\frac{\sqrt{3}}{4}\right)}^2=\pi \cdot \frac{9}{16}=\frac{9}{16}\pi \left[mm\right]$  

$Obw=2\cdot \pi \cdot \frac{\sqrt{3}}{4}=2\cdot \frac{\sqrt{3}}{4}\cdot \pi =\frac{2\sqrt{3}}{4}\pi \left[mm\right]$

 

 

$d)d=\pi cm$  

$\text{a więc}r=\frac{\pi }{2}$

$P=\pi \cdot {\left(\frac{\pi }{2}\right)}^2=\pi \cdot \frac{{\pi }^2}{4}=\frac{\pi \cdot {\pi }^2}{4}=\frac{{\pi }^3}{4}\left[cm\right]$

$Obw=2\cdot \pi \cdot \frac{\pi }{2}=\frac{2\cdot \pi \cdot \pi }{2}=\frac{2\cdot {\pi }^2}{2}=\frac{{\sout{2}}^1\cdot {\pi }^2}{{\sout{2}}^1})={\pi }^2\left[cm\right]$

 

 

$e)d=1,6dm$  

$\text{a więc}r=0,8dm$

$P=\pi \cdot {\left(0,8\right)}^2=\pi \cdot 0,64=0,64\pi \left[dm\right]$

$Obw=2\cdot \pi \cdot 0,8=2\cdot 0,8\cdot \pi =1,6\pi \left[dm\right]$

 

 

$f)d=3\sqrt{2}cm$  

$\text{a więc}r=\frac{3\sqrt{2}}{2}cm$  

$P=\pi \cdot {\left(\frac{3\sqrt{2}}{2}\right)}^2=\pi \cdot \frac{9\cdot 2}{4}=\frac{\pi \cdot 9\cdot {\sout{2}}^1}{{\sout{4}}^2}=\frac{9}{2}\pi \left[cm\right]$  

$Obw=2\cdot \pi \cdot \frac{3\sqrt{2}}{2}=2\cdot \frac{3\sqrt{2}}{2}\cdot \pi ={\sout{2}}^1\cdot \frac{3\sqrt{2}}{{\sout{2}}^1}\cdot \pi =3\sqrt{2}\pi \left[cm\right]$