Odpowiedź :
[tex]a = 3 \ cm\\b = \sqrt{7} \ cm\\c = ?\\Ob = ?[/tex]
Z tw. Pitagorasa liczę przeciwprostokątną c:
[tex]a^{2}+b^{2} = c^{2}\\\\3^{2} + (\sqrt{7})^{2} = c^{2}\\\\9+7 = c^{2}\\\\c^{2} = 16\\\\c = \sqrt{16}\\\\\underline{c = 4 \ cm}[/tex]
[tex]Ob = a+b+c\\\\Ob = 3 \ cm + \sqrt{7} \ cm + 4 \ cm\\\\\boxed{Ob = (\sqrt{7}+7) \ cm}\\\\Odp. \ A.[/tex]
[tex]a^2+b^2=c^2\\a=3cm\\b=\sqrt7cm\\\\(3cm)^2+(\sqrt7cm)^2=c^2\\9cm^2+7cm^2=c^2\\16cm^2=c^2\\c=4cm\\\\Ob=a+b+c\\Ob=3cm+\sqrt7cm+4cm=(7+\sqrt7)cm\\\\Odp. A[/tex]