in tr. dreptunghic ADB, ∡B=45, ⇒∡BAD=45 ⇒ AD=BD=6√3
cu pitagora in ADC
AC^2=AD^2+DC^2=AD^2+AC^2/4 (∡DAC=30° ⇒ DC=AC/2, teorema ∡30°)
AC^2=3 x 36+AC^2/4
3AC^2=12 x 36
AC=12
AB=AD√2 (pitagora in tr. dreptunghic isoscel ADB)
AB=6√6
BC=BD+DC=6√3+6
BC=6(1+√3)
aria tr. ABC, At
At=BC x AD/2=6(1+√3) x 6√3/2
At=18√3(1+√3)
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