Z twierdzenia Pitagorasa:
[tex]a^{2}+b^{2}=c^{2}[/tex]
b)
[tex](3\sqrt{5})^{2}+b^{2}=(4\sqrt{3}) ^{2}\\45+b^{2}=48\\b^{2} =48-45\\b^{2} =3\\b=\sqrt{3}[/tex]
"a" i "b" to przyprostokątne, a "c" to przeciwprostokątna
c)
[tex](4\sqrt{3}) ^{2} +b^{2} = 7^{2} \\48+b^{2} = 49\\b^{2} = 49-48\\b^{2}= 1\\b=\sqrt{1} = 1[/tex]
