Zad. 1
[tex]L=2\pi r = \pi d[/tex]
[tex]a)\\r=0.4cm\\L=2\pi * 0.4cm=0.8\pi cm\\\\b) \\d=2m\\L=2\pi m[/tex]
[tex]c)\\L=628mm\\2\pi r=628mm /:2\\\pi r=314mm /:pi\\r=\frac{314}{\pi}mm = ok. 100mm[/tex]
Zad. 3
[tex]d_k = 8m\\r_k=\frac{8m}2=4m\\P_k=16\pi m^2\\\\r_{k+c}=4m+2m=6m\\P_{k+c}=36\pi m^2\\\\P_c=P_{k+c}-P_k\\P_c=36\pi m^2-16\pi m^2=20\pi m^2=ok. 62.8m^2\\\\\text{Chodnik ma powierzchnie } 20\pi m^2 \text{ czyli okolo }62.8m^2[/tex]
Zad. 4
[tex]r=4m\\L=2*4m*\pi = 8\pi m\\5L=5*8\pi m=40\pi m=ok. 125.6m\\\\\text{Kon pokonal droge o dlugosci }40\pi m \text{ czyli okolo } 125.6m[/tex]
[tex]10km=10000m[/tex]
[tex]\text{1 okrazenie: } 8\pi m = 25.12m\\\frac{10000m}{25.12m}=398.089...\\399*8\pi m > 10km\\\\\text{Kon musial przebiec 399 okrazen}[/tex]
