Odpowiedź:
c) [tex]3^{2} + (3\sqrt{7})^{2}[/tex] = [tex]x^{2}[/tex]
9 + 9*7 = [tex]x^{2}[/tex]
9 +63 = [tex]x^{2}[/tex]
[tex]x^{2}[/tex] = 72
x = [tex]\sqrt{72}[/tex]
x = 6 [tex]\sqrt{2}[/tex]
d) [tex]6^{2} +(2\sqrt{3})^{2} = x^{2}[/tex]
[tex]36 + 4* 3 = x^{2} \\36 + 12 = x^{2} \\48 = x^{2} \\x= \sqrt{48} \\x = 4 \sqrt{3}[/tex]
e)
[tex]6^{2} + 4^{2} = x^{2} \\36 + 16 = x^{2} \\52 = x^{2} \\x = 2 \sqrt{13}[/tex]
f)
[tex](2\sqrt{5} )^{2} + x^{2} = (2\sqrt{19} )^{2} \\4* 5 + x^{2} = 4*19 \\20 + x^{2} = 76 \\x^{2} = 56\\[/tex]
x = 2[tex]\sqrt{14}[/tex]
Szczegółowe wyjaśnienie:
Zadania wykonujemy zgodnie z twierdzeniem Pitagorasa
wdg wzoru [tex]a^{2}[/tex] + [tex]b^{2}[/tex] = [tex]c^{2}[/tex]
