[tex]P_{troj}=3\sqrt{3}\\\frac{a^2\sqrt{3}}{4}=3\sqrt{3}|:\sqrt{3}\\\frac{a^2}{4}=3|*4\\a^2=12|\sqrt{~}\\a=\sqrt{12}\\a=2\sqrt{3}\\\\h=\frac{a\sqrt{3}}{2}\\h=\frac{2\sqrt{3}*\sqrt{3}}{2}\\h=3\\\\S_{podst}=\sqrt{6}\\r_{podst}=\frac{\sqrt{6}}{2}\\\\V=\frac{1}{3}P_p*h\\V=\frac{1}{3}\pi r^2*h\\V=\frac{1}{3}\pi (\frac{\sqrt{6}}{2})^2*3=\pi \frac{6}{4}=1\frac{1}{2}\pi[/tex]
Odp. Objętość stożka wynosi [tex]1\frac{1}{2}\pi[/tex].
