Odpowiedź:
twierdzenie pitagorasa
[tex] {a}^{2} + {b}^{2} = {c}^{2} [/tex]
"a" i "b" są to przyprostokątne, a "c" przeciwprostokątna
POZIOM A
[tex]a) {2}^{2} + {5}^{2} = {x}^{2} \\ 4 + 25 = {x}^{2} \\ {x}^{2} = 29 \\ x = \sqrt{29} [/tex]
[tex]b) {1}^{2} + {x}^{2} = {2}^{2} \\ 1 + {x}^{2} = 4 \\ {x}^{2} = 4 - 1 \\ {x}^{2} = 3 \\ x = \sqrt{3} [/tex]
POZIOM B
[tex]a) {2}^{2} + ( \sqrt{21} {)}^{2} = {x}^{2} \\ 4 + 21 = {x}^{2} \\ {x}^{2} = 25 \\ x = 5[/tex]
[tex]b) (\sqrt{3} {)}^{2} + {x}^{2} = (\sqrt{5} {)}^{2} \\ 3 + {x}^{2} = 5 \\ {x}^{2} = 5 - 3 \\ {x}^{2} = 2 \\ x = \sqrt{2} [/tex]
