Fie H∈(BC) astfel incat AH _|_ BC => H-mijlocul segm. [BC] => BH=CH=6cm.
Aplic teorema lui Pitagora in ΔABH-dreptunghic in H:
[tex] AH^{2}+ BH^{2}= AB^{2} => AH= \sqrt{ AB^{2}- BH^{2} } = \sqrt{100-36}= \sqrt{64}= \\ =8(cm). \\ \\ A_{ABC}= \frac{AH*BC}{2}= \frac{8*12}{2}=48( cm^{2}) [/tex]
