Zastosujmy twierdzenie Pitagorasa
[tex] {5}^{2} + {9}^{2} = {x}^{2} \\ 25 + 81 = {x}^{2} \\ 106 = {x}^{2} \\ x = \sqrt{106} cm[/tex]
[tex]a = 5 \ cm\\b = 9 \ cm\\d = ?\\\\Z \ tw. \ Pitagorasa:\\\\d^{2} = a^{2}+b^{2}\\\\d^{2} = 5^{2}+9^{2}\\\\d^{2} = 25+81\\\\d^{2} = 106\\\\d = \sqrt{106} \ cm\\\\\boxed{d \approx10,3 \ cm}[/tex]
