Răspuns :
Δ ABC-dr⇒conform T.P: AC²=BC²-AB²=12²-(6√3)²=6²(4-3)=6²⇒AC=6
OA=AC=AC=6⇒ΔOAC-echilateral⇒m<AOC=60
A hasurata=A sector-A ΔOAC=π·R²·n/360-l²√3/4=π36·60/360-36√3/4=
=6π-9√3=3(2π-3√3)
OA=AC=AC=6⇒ΔOAC-echilateral⇒m<AOC=60
A hasurata=A sector-A ΔOAC=π·R²·n/360-l²√3/4=π36·60/360-36√3/4=
=6π-9√3=3(2π-3√3)