Question

1.Triunghiul ABC are AB =12 cm , AC=15 cm si BC=18. Prin punctul D€(AB) cu AD=8 se duce paralela DE la BC (E€(AC)).Aflati perimetrul triunghiului ADE

2.Trapezul ABCD are bazele AB =36 cm, CD=21 cm, iar laturile neparalele AD=20 cm si BC=25cm.Daca AD ù BC={P}, calculati PA, PB, PC, PD.

Va rog frumos ajutatima!!

Answer 1

1)
DE ||BC ⇒ ΔADE asemeneaΔ ABC
⇒AD/AB=AE/AC=DE/BC⇒ 8/12=AE/15=DE/18
AE= 8*15:12=10 cm
DE= 8*18:12=12 cm
P ADE=AD+DE+AE=8+12+10=30 cm

2)
CD||AB⇒ ΔPAB asemenea Δ PDC ⇒ 
⇒PD/PA=PC/PB=CD/AB ⇒ PD/PA=PC/PB=21/36=7/12
PD/PA=7/12
PD/PA-PD = 7/12-7⇒ PD/20= 7/5⇒ PD=20*7:5=28 cm
PA=AD+PD=20+28=48 cm
PC/PB=7/12⇒ PC/PB-PC=7/12-7 ⇒ PC/25=7/5⇒ PC=25*7:5=35cm
PB=PC+BC=35+25=60cm

Answer 2

Rezolvare Probl.1 si Probl.2 ................