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Răspuns :

a)
aria = cateta 1*cateta2/2=AB*AC/2=6*6/2=18m²

b) 2*πR²/2=πR² unde R=6/2=3m
deci aria hasurata =π*3²=9π m²

c)d (A, BC) =inaltimea inΔABC= mediana in tr.dr.isoscelABC=BC/2= 6√2/2=3√2 m

as simple as that!