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AD=36
AC=60
ΔABC
AD⊥BC
m(<A)=90
D∈(BC)
Aflati BD,CD,AB,BC


Răspuns :

In dreptunghic ΔADC
AC²=DC²+AD²
CD²=AC²-AD²=60²-36²=3600-2396=2304
CD=48

Th inaltimii:
AD²=CD*DB
DB=AD²/CD=36²/48⇒DB=27

BC=BD+DC=27+48⇒BC=75

BC²=AB²+AC²
=>AB²=BC²-AC²=75²-60²=5625-3600=2025
AB=45
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