Răspuns :

Ai rezolvarea în atașament. Mult succes!

Vezi imaginea Loredanaschneid

[tex]\it a)\\ \\ \left.\begin{aligned}VABCD\ -\ piramid\breve a\ patrulater\breve a\ regulat\breve a\\ \\ O\ -\ centrul\ cercului\ circumscris\ bazei \end{aligned}\right\} \Rightarrow\ VO- \^{i}n\breve al\c{\it t}imea\ piramidei\Rightarrow[/tex]

[tex]\it \Rightarrow VO\perp (ABC)\ \c{s}i\ BO\subset (ABC) \Rightarrow VO\perp BO \Rightarrow BO\perp VO\ \ \ \ \ \ (1)\\ \\ BO\perp AC\ \ \ \ \ \ (2)\\ \\ VO,\ AC\ \subset (VAC)\ \ \ \ \ \ (3)\\ \\ (1),\ (2),\ (3) \Rightarrow BO\perp (VAC) \Rightarrow d[B,\ (VAC)]=BO[/tex]

[tex]\it BO=\dfrac{BD}{2}=\dfrac{\ell \sqrt2}{2}=\dfrac{6\sqrt2\csot\sqrt2}{2}=\dfrac{6\cdot2}{2}=6\ cm[/tex]

[tex]\it b)\\ \\ AV=CV=6\sqrt2\ cm\\ \\ AC=BD=2\cdot BO=2\cdot6=12\ cm\\ \\ AV^2+CV^2=(6\sqrt2)^2+(6\sqrt2)^2=36\cdot2+36\cdot2=144=12^2=AC^2\Rightarrow\\ \\ \Rightarrow \Delta CVA-dr, \hat V=90^o\ (din\ reciproca\ teoremei\ lui\ Pitagora) \\ \\ Deci,\ \ CV\perp AV \Rightarrow (\widehat{AV,\ CV})=90^o[/tex]