Question

fie abcd un dreptunghi cu ab=4cm si bc = 2 cm. in punctul b se construieste perpendiculara pe planul dreptunghiului, pe care se considera punctul E astfel incat EB= 2 radical 2. Calculati:a)d(E,AD);b)d(E,CD)

Answer 1

Răspuns:

EB perpendicular pe (ABC) BC perpendicular pe DC BC, Dc incluse in planul ( ABC) => conform teoremei celor trei perpemdiculare ca EC este perpendicular pe DC => d ( E, CD ) = EC IN TR EBC = triunghi dreptunghic in EBC => conform teoremei lui pitagora ca EC LA PATRAT = 2 LA A DOUA + 2 LA A DOUA = 8 => EC =RADICAL DIN 8 = 2 RADICAL DIN 2

Answer 2

a)

$\it \left.\begin{aligned} \it EB\perp(ABC)\\ \\ \it BA\perp AD\ \ \ \ \\ \\ BA,\ AD\subset (ABC) \end{aligned}\right\} \stackrel{T3\perp}{\Longrightarrow}\ EA\perp AD \Rightarrow d(E,\ AD)=EA\\ \\ \\ EB\perp(ABC),\ \ BA\subset(ABC) \Rightarrow EB\perp BA \Rightarrow \Delta EBA-dreptunghic,\\ \\ m(\widehat{EBA})=90^o \stackrel{TP}{\Longrightarrow}\ EA^2=EB^2+BA^2=(2\sqrt2)^2+4^2=8+16=24 \\ \\ EA=\sqrt{24}=\sqrt{4\cdot6}=2\sqrt6\ cm \Rightarrow d(E,\ AD)=2\sqrt6\ cm$

b)

$\it \left.\begin{aligned} \it EB\perp(ABC)\\ \\ \it BC\perp CD\ \ \ \ \\ \\ BC,\ CD\subset (ABC) \end{aligned}\right\} \stackrel{T3\perp}{\Longrightarrow}\ EC\perp CD \Rightarrow d(E,\ CD)=EC\\ \\ \\ EB\perp(ABC),\ \ BC\subset(ABC) \Rightarrow EB\perp BC \Rightarrow \Delta EBC-dreptunghic,\\ \\ m(\widehat{EBC})=90^o \stackrel{TP}{\Longrightarrow}\ EC^2=EB^2+BC^2=(2\sqrt2)^2+2^2=8+4=12\\ \\ EA=\sqrt{12}=\sqrt{4\cdot3}=2\sqrt3\ cm \Rightarrow d(E,\ CD)=2\sqrt3\ cm$