d=6\sqrt{3}
\alpha=30st.
V=P_pH=a^2H
\sin\alpha=\frac{H}{d}
\sin30=\frac{H}{6\sqrt{3}}
\frac{1}{2}=\frac{H}{6\sqrt{3}}
H=\frac{H}{6\sqrt{3}}*\frac{1}{2}=3\sqrt{3}
c - przekątna podstawy (kwadratu)
c^2=d^2-H^2=(6\sqrt{3})^2-(3\sqrt{3})=108-27=81
c=\sqrt{81}=9
c=a\sqrt{2}
a=\frac{c}{\sqrt{2}}=\frac{9}{\sqrt{2}}=\frac{9*\sqrt{2}}{\sqrt{2}*\sqrt{2}}=\frac{9\sqrt{2}}{2}=4,5\sqrt{2}
V=a^2H=(4,5\sqrt{2})^2*3\sqrt{3}=121,5\sqrt{3}cm^3 co można zapisać V=\frac{243\sqrt{3}}{2}cm^3