Question

help pleaseeeeeeeeeee
Dreptunghiul Abcd ,Ab=18 radical din 3 ,Ad=18,AE perpendicular BD , E apartine BD ,AE intersectat DC (F) a) Aflati AE , AF , DF , Perimetrul , Aria dreptunghiului
b)Aria DEF / Aria AEB

Answer 1

[tex]DB^2=AD^2+AB^2 \\ DB^2=324+972\\ DB^2=1296[/tex]⇒$DB=36 cm$
[tex]AE= \frac{AB*AD}{BD}\\ AE=\frac{18*18 \sqrt{3}}{36}\\ AE=9\sqrt{3} cm\\ DE^{2}=AD^2-AE^2\\ DE^2=324-243\\ DE^2=81 [/tex]⇒DE= 9 cm
ΔDBC:∡D comun
ΔDFE:∡DEF=∡DCB
Din ambele rezulta prin UU ca ΔDBC≈ΔDEF⇒[tex] \frac{EF}{BC}= \frac{DF}{DB}= \frac{DE}{CD}\\ \frac{EF}{18}= \frac{DF}{36}= \frac{9}{18 \sqrt{3} }\\ DF= \frac{36*9}{18 \sqrt{3} }=6 \sqrt{3}cm\\ EF= \frac{18*9}{18 \sqrt{3} }=3 \sqrt{3}cm [/tex]
AF=AE+EF
AF=9√3+3√3=12√3 cm
b)EB=DB-DE
EB=36-9=27 cm
[tex] A_{AEB}= \frac{AE*EB}{2}\\ A_{AEB}= \frac{9 \sqrt{3}*27 }{2}= \frac{243 \sqrt{3} }{2}cm^2\\ A_{EDF}= \frac{EF*DE}{2}\\ A_{EDF}= \frac{9* 3\sqrt{3} }{2}= \frac{27 \sqrt{3} }{2}cm^2\\ \frac{A_{AEB}}{A_{EDF}}= \frac{ \frac{243 \sqrt{3} }{2} }{ \frac{27 \sqrt{3} }{2} }\\ \frac{A_{AEB}}{A_{EDF}}= \frac{243 \sqrt{3} }{2}* \frac{2}{27 \sqrt{3} }\\ \frac{A_{AEB}}{A_{EDF}}=9 [/tex]