Question

Problema 8....va rog mult
URGENT!!
50 PCT

Answer 1

1)AD=4dam
   AE=4AB/5=8dam
DE=√(4²+8²)=4√5dam ( Teo Pitagora in tr.dr.DAE)
PerimΔADE=4+8+4√5=(12+4√5)dam


BC=AD=4
BE=AB/5=10/5=2
BE=√(2²+4²)=2√5 (Teo Pitagora in tr.dr,.CEB)
PerimΔCEB=2+4+2√5=6+2√5

b) AM med in tr dr de ipotenuza 4√5, AM=2√5
     MN l.m.tr DEC, MN=10:2=5
     NB med in tr.fr de ipotenuza 2√5, MN=√5
 Lung dru,=AM+MN+NB=2√5+5+√5=5+3√5
c) var 1
DE²=(4√5)²=80
EC²=(2√5)²=20
DC²=10²=100
DE²+EC²=DC²⇒(Rec. Teo Pitagora) ΔDEC drptunghic in e, mas∡DEC=90°

var 2 altfel
AD/EB=4/2=2
AE/BC=8/4=2
mas ∡A=mas∡B=90°, caz LUL (2 perechi lat prop, 2 ungh congr),  ΔDAE≈ΔEBC⇒
∡AED≡∡ECB
dar mas DEC=180°-mas∡AED-mas∡CEB
 deci mas DEC =180°-mas∡ECB-mas∡CEB*=180°-90°=90°
 
(*)∡ECB si ∡CEB, complementare, Δ CEB, dreptunghic