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Se considera triunghiul ascutitunghic ABC, in care AB=26 cm, BC =42 cm,AD perpendicular BC,D apartine BC si AD=24 CM.Perpendiculara in A pe AB intersecteaza dreapta BC in E. Aflati perimetrul Δ ACE. 

Răspuns :

MFMid
DB²=AB²-AD²=26²-24²=676-576=100
DB=10⇒ CD=32
AC²=AD²+CD²=24²+32²=576+1024=1600
AC=40 CM

punctu E se afla in afara tr ABC deoarece EAC dreptunghic verificam th inaltimii
AD²=DB.DE
576=10.ED
ED=57,6 > CD =32

din ΔADE, AE²=AD²+ED²
AE²=576+576.576/100=576(100+576)/100
AE=24.26/10=624/10

EC=ED-DC=576/10-32=
256/10
P=256/10+624/10+400/10=1280/10=128 cm

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