[tex]Zadanie\\\\a=6\\c=9\\\\Przekatna\ podstawy\\\\d=a\sqrt{2}\\\\d=6\sqrt{2}\\\\ Korzystamy\ z\ Twierdzenia\ Pitagorasa\\\\(\frac{d}{2})^2+x^2=c^2\\\\ (\frac{6\sqrt{2} }{2})^2+x^{2}=9^2\\\\(3\sqrt{2})^2+x^{2}=81\\\\18+x^{2}=81\ \ \mid-18 \\\\x^{2}=63\\\\x=\sqrt{63}=\sqrt{9\cdot7}\\\\x=3\sqrt{7}\ (j)[/tex]
