$P_p={\left(6\text{cm}\right)}^2=36{\text{cm}}^2$  

 

$d=a\sqrt{2}=6\sqrt{2}\text{cm}$  

$\frac{1}{2}d=\frac{1}{2}\cdot 6\sqrt{2}\text{cm}=3\sqrt{2}\text{cm}$  

${\left(3\sqrt{2}\text{cm}\right)}^2+H^2={\left(6\text{cm}\right)}^2$  

$3^2\cdot {\left(\sqrt{2}\right)}^2{\text{cm}}^2+H^2=36{\text{cm}}^2$  

$9\cdot 2{\text{cm}}^2+H^2=36{\text{cm}}^2$  

$18{\text{cm}}^2+H^2=36{\text{cm}}^2\mid -18{\text{cm}}^2$  

$H^2=18{\text{cm}}^2\mid \sqrt{}$  

$H=\sqrt{18}\text{cm}=\sqrt{9\cdot 2}\text{cm}=3\sqrt{2}\text{cm}$  

 

$V=\frac{1}{3}\cdot P_p\cdot H=\frac{1}{{\sout{3}}^1}\cdot 36{\text{cm}}^2\cdot {\sout{3}}^1\sqrt{2}\text{cm}=36\sqrt{2}{\text{cm}}^3$    

Odpowiedź:

Objętość tego ostrosłupa wynosi 36√2 cm³.