3. Zadatak

Vrijedi sljedeće:

$$P_1=9\pi,\quad \frac{r_1^2\pi}{2}=9\pi,\quad r_1=\sqrt{18}=3\sqrt2,\quad a=6\sqrt2,$$ $$P_2=16\pi,\quad \frac{r_2^2\pi}{2}=16\pi,\quad r_2=\sqrt{32}=4\sqrt2,\quad b=8\sqrt2,$$ $$P_3=25\pi,\quad \frac{r_3^2\pi}{2}=25\pi,\quad r_3=\sqrt{50}=5\sqrt2,\quad c=10\sqrt2.$$

Iz Heronove formule slijedi

$$s=12\sqrt2,\quad P=\sqrt{12\sqrt2\cdot6\sqrt2\cdot4\sqrt2\cdot2\sqrt2}=\sqrt{2304}=48.$$

Napomena: Uočimo li da je trokut pravokutan ($a^2+b^2=72+128=200=(10\sqrt2)^2=c^2$), površinu možemo računati po formuli $P=\frac{ab}{2}=48$.