Question

Problema 13 si 16 (poza modificata) 100 PCT
+COROANA
URGENT!!

Answer 1

DE=6 sin60°=6*√3/2=3√3cm
AE=6cos60°=6*1/2=3cm

b)EB=8-3=5
DE=3√3
BD=(PitagoraΔDEB)=√((3√3)²+5²)=√(27+25)=√52=2√13

c) fie CF⊥AB, F∈AB⇒BF=3, CF=3√3 (ΔAED≡ΔBFC, caz CU)
AC²=AF²+CF²
AC²=(8+3)²+(3√3)²
AC²=121+27
AC²=148
AC=√148=2√39

16

a) ABC, tr dr isoscel, BD ipotenuza⇒m∡ABD=45°
DBC echilkatteral ⇒m∡DBC=60°
deci m∡ABC=45°+60°=105°

b)AB=BDcos45°=60/√2=30√2
30√2 comparat 45 simplificam cu 15
2√2 comparat cu 3
2√2<3 pt ca, ridicand la patrat
8<9

c)fie AC∩BD={O}
AC=AO+OC
AD≡AB ( ADB isoscel) ⇒D ∈medioatoarei segment BD (1)
CD≡CB (CDB echilateral)⇒C∈medioatoarei segment BD (2)
din (1) si (2)⇒AC⊥DB⇒AO si CO inaltimi in ΔADB si, respectiov, ΔBCD
AO inaltime , ADB isoscelAO mediana, AO=BD/2=60/2=30
CO inaltyime, CBD echilateral CO=60√3/2=30√3
DB=60 (latura tr.echilateral, ipoteza)
total bagheta ;30+30√3+60=90+30√3≈141,96..≈142cm