Z twierdzenia Pitagorasa:
a² + b² = c²
gdzie:
a,b - przyprostokątne
c - przeciwprostokątna
a)
[tex]x^{2}+2^{2} = \sqrt{24}^{2}\\\\x^{2} + 4 = 24\\\\x^{2} = 24-4 = 20\\\\x = \sqrt{20} = \sqrt{4\cdot5}\\\\\boxed{x = 2\sqrt{5}}[/tex]
b)
[tex]5^{2}+x^{2} = 7^{2}\\\\25+x^{2} = 49\\\\x^{2} = 49-25 = 24\\\\x = \sqrt{24} = \sqrt{4\cdot6}\\\\\boxed{x = 2\sqrt{6}}[/tex]
c)
[tex](3\sqrt{7})^{2}+3^{2} = x^{2}\\\\63+9 = x^{2}\\\\x^{2} = 72\\\\x = \sqrt{72} = \sqrt{36\cdot2}\\\\\boxed{x = 6\sqrt{2}}[/tex]
d)
[tex]6^{2}+(2\sqrt{3})^{2} = x^{2}\\\\36+12=x^{2}\\\\x^{2} = 48\\\\x = \sqrt{48} = \sqrt{16\cdot3}\\\\\boxed{x = 4\sqrt{3}}[/tex]
e)
[tex]6^{2}+4^{2} = x^{2}\\\\36+16 = x^{2}\\\\x^{2} = 52\\\\x = \sqrt{52}\\\\x = \sqrt{4\cdot13}\\\\\boxed{x = 2\sqrt{13}}[/tex]
f)
[tex]x^{2} + (2\sqrt{5})^{2} = (2\sqrt{19})^{2}\\\\x^{2}+20=76\\\\x^{2} = 76-20 = 56\\\\x = \sqrt{56} = \sqrt{4\cdot14}\\\\\boxed{x = 2\sqrt{14}}[/tex]
