a = 8cm
c = 12 cm
b^2=c^2-a^2
{b=\sqrt{a^2+b^2}=\sqrt{12^2-8^2}=\sqrt{144-64}=\sqrt{80}=\sqrt{16\cdot 5}=4\sqrt{5} \ (cm)}
a)
r=\frac{a+b-c}{2}=\frac{8+4\sqrt5-12}{2}=\frac{4\sqrt5-4}{2}=2\sqrt5-2
{P_{kola}=\pi r^2=\pi \cdot [2(\sqrt5-1)]^2=4\pi \cdot(5-2\sqrt5+1)=4\pi (6-2\sqrt5)=8\pi (3-\sqrt5) \ (cm^2)} <-- odpowiedź
\approx 8*3.14(3-2.236)\approx 19,2
b)
Pole trójkąta
P=\frac{a\cdot b}{2}=\frac{8\cdot 4\sqrt5}{2}=16\sqrt5\ (cm^2) <-- odpowiedź