Question

Ajutor,am nevoie de rezolvare,urgent !

Answer 1

a)
mas ∡EBD=180°-60°-30°=90°

b)AriaΔ EBD=EB*BD/2
EB=6
BD=(18/2):(√3/2)=18:√3=6√3

Aria Δ EBD=6*6√3/2=18√3 cm²


c)
inΔEBD, EB=6, BD=6√3,m∡EBD=90°⇒(Teo Pitagora)⇒ED=12; cum EB=6⇒m∡EDB=30°⇒m∡BED=60°(complementar ΔEDB) dar si m∡EBA=60°, dar alterme interne drepte ED,AB si BE secanta⇒ED||AB, C.C.T.D.