Question

Fie piramida hexagonală regulată VABCDEF, V fiind vârful piramidei. Dacă tetraedrul VEAC este regulfat iar AB = 6 cm, deteminați lungimea înălțimii unei fețe laterale ale piramidei VABCDEF.​

Answer 1

$\it VEAC-tetraedru\ regulat \Rightarrow VB= VA=AC\ \ \ \ \ (1)\\ \\ Fie\ BQ- \hat\imath n\brebe al\c{\it t}ime\ pentru\ \Delta ABC, unde\ BC=AB=6\ cm,\ \widehat{ABC}=120^o\\ \\ \widehat{BCA}=\widehat{CAB}=(180^o-120^0):2=30^o \Rightarrow \Delta QBC\ are\ forma\ (30^o,\ 60^o,\ 90^o),\\ \\ cu\ ipotenuza\ BC=6\ cm \Rightarrow BQ=3\ cm,\ QC=3\sqrt3\ cm,\\ \\ AC=2\cdot3\sqrt3=6\sqrt3cm\ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow VB=VA=6\sqrt3cm$

$\it Fie\ VM\ -\ \hat\imath n\breve a l\c{\it t}ime\ pentru\ fa\c{\it t}a\ lateral\breve a\ VAB \Rightarrow VM\ este\ \d si\ median\breve a \Rightarrow \\ \\ \Rightarrow AM=MB=6:2=3\ cm\\ \\ \Delta VMA- dreptunghic, \ \widehat{M}=90^o,\ \stackrel{T.P.}{\Longrightarrow}\ VM^2=VA^2-AM^2=(6\sqrt3)^2-3^2=\\ \\ =36\cdot3-9=9(4\cdot3-1)=9\cdot11 \Rightarrow VM=\sqrt{9\cdot11}=3\sqrt{11}\ cm$