Wszystkie trójkąty są prostokątne, więc korzystamy z twierdzenia Pitagorasa:
a² + b² = c²
gdzie:
a,b - długości przyprostokątnych
c - długość przeciwprostokątnej
a)
[tex]x^{2} = 2^{2}+\sqrt{21}^{2}\\\\x^{2} = 4 + 21\\\\x^{2} = 25\\\\x = \sqrt{25}\\\\\boxed{x = 5}[/tex]
b)
[tex]x^{2} = \sqrt{5}^{2} - \sqrt{3}^{2}\\\\x^{2} = 5-3\\\\x^{2} = 2\\\\\boxed{x = \sqrt{2}}[/tex]
c)
[tex]x^{2} = \sqrt{21}^{2} -(2\sqrt{5})^{2}\\\\x^{2} = 21 - 20\\\\x^{2} = 1\\\\\boxed{x = 1}[/tex]
d)
[tex]x^{2} = \sqrt{8}^{2}+3^{2}\\\\x^{2} = 8+9\\\\x^{2} = 17\\\\\boxed{x = \sqrt{17}}[/tex]
e)
[tex]x^{2} = (2\sqrt{3})^{2}+2^{2}\\\\x^{2} = 12+4\\\\x^{2} = 16\\\\x = \sqrt{16}\\\\\boxed{x = 4}[/tex]
f)
[tex]x^{2}= 7^{2}-(3\sqrt{5})^{2}\\\\x^{2} = 49-45\\\\x^{2} = 4\\\\x = \sqrt{4}\\\\\boxed{x = 2}[/tex]
