[tex]objetosc\ ostroslupa\ prawidlowego\ czworokatnego : V=36\sqrt{2}\ cm^3 \\\\krawedz:\ a=?\\\\V=\frac{1}{3}P_{p}*H\\\\ obliczamy\ wysokosc\ graniastosludpa\ stosujac\ tw.\ Pitagorasa:\\\\przekatna\ podstawy:\\\\d=a\sqrt{2}\\\\ H^2+(\frac{d}{2})^2=a^2\\\\H^2=a^2-(\frac{d}{2})^2[/tex]
[tex]H^2=a^2-(\frac{a\sqrt{2}}{2})^2 \\\\H^2=a^2- \frac{2a^2}{4} \\\\H^2=a^2- \frac{ 1}{2}a^2\\\\H^2= \frac{ 1}{2}a^2\\\\H =\sqrt{ \frac{ 1}{2}a^2}= \frac{a }{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}} =\frac{\sqrt{2}a}{2}\ cm\\\\pole\ podstawy:\\\\P_{p}=a^2[/tex]
[tex]\frac{1}{3}*a^2 *\frac{\sqrt{2}\ a}{2}=36\sqrt{2} \ \ |*\sqrt{2}\\\\\frac{1}{3}*\frac{2a^3}{2}=36*2\\\\\frac{1}{3}a^3=72\ \ |*3\\\\a^3=216\\\\a=\sqrt[3]{216} \\\\a=6\ cm\\\\odp.\ Krawedzie\ tego\ ostroslupa\ maja\ po\ 6\ cm\ dlugosci.[/tex]
