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Dosabogdan11
a fost răspuns

Trapezul ABCD are baza mare AB SI AB ll DC , [AD] ≡ [DC] , AD = 6 cm BC = 3√5 , iar d(DC , AB )= 2√5 . Sa se afle lungimile segmentelor AB , AC si DB .


Răspuns :

Ducem DM_|_AB si CN_|_AB=> d(DC , AB )= DM=CN=2√5 si DC=MN
in ΔDMA:
AM²=AD²-DM²=36-20=16
AM=4
InΔCNB avem
NB²=CB²-CN²=45-20=25
NB=5
Daca AD=DC=6
=>AB=AM+MN+NB=4+6+5
AB=15

in ΔACN
AC²=AN²+CN²=100+20=120
AC=2√30

in ΔDMB
DB²=DM²+MB²=20+121=141
DB=√141

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