[tex]\displaystyle\bf\\1)\\\\a)\\Se~da:~\triangle ABC~cu m(\angle A)=90^o,~AB=7~cm,~AC=7\sqrt{2}~cm\\Se~cere:~BC=?\\Rezolvare\!:\\BC=\sqrt{AB^2+AC^2}=\sqrt{7^2+(7\sqrt{2})^2}=\sqrt{7^2+7^2\cdot2}=\\\\=\sqrt{7^2+7^2\cdot2}=\sqrt{7^2(1+2)}=7\sqrt{3}~cm\\\\\\ b)\\Se~da:~\triangle ABC~cu~m(\angle A)=90^o,~AC=2\sqrt{5}~cm,~BC=5~cm,\\Se~cere:~AB=?\\Rezolvare\!:\\AB=\sqrt{BC^2-AC^2}=\sqrt{5^2-(2\sqrt{5})^2}=\sqrt{5^2+2^2\cdot5}=\sqrt{25+20}=\sqrt{5}~cm[/tex]
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[tex]\displaystyle\bf\\c)\\Se~da:~\triangle ABC~cu~m(\angle A)=90^o,~m(\angle C)=30^o~si~BC=8~cm,\\Se~cere:~AB=?~si~AC=?\\Rezolvare\!:\\AB=\frac{BC}{2}=\frac{8}{2}=4~cm~(fiind~opusa~unghiului~C~de~30^o)\\AC=\sqrt{BC^2-AB^2}=\sqrt{8^2-4^2}=\sqrt{64-16}=\sqrt{48} =\sqrt{16\cdot3}=4\sqrt{3}[/tex]
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[tex]\displaystyle\bf\\d)\\Se~da\!:~\triangle ABC~cu~m(\angle A)=90^o,m(\angle C)=60^o,\\~~~~~~~~~~~~M=mijl.~BC~si~AM=3\sqrt{2}~cm\\Se~cere:~P \triangle ABC=?\\Rezolvare:\\AM=mediana~lui~BC.\\\implies~BC=2 \cdot AM=2 \cdot 3\sqrt{2}=6\sqrt{2}~cm\\m(\angle B)=60^o=90^o-60^o=30^o\\AC=\frac{BC}{2}=\frac{6\sqrt{2}}{2}=3\sqrt{2}~cm\\AB=\sqrt{BC^2-AC^2}=\sqrt{(6\sqrt{2})^2-(3\sqrt{2})^2}=\\=\sqrt{6^2\cdot2-9\cdot2}=\sqrt{72-18}=\sqrt{54}=3\sqrt{6}\\P=3\sqrt{6}+3\sqrt{2}+6\sqrt{2}=3\sqrt{6}+9\sqrt{2}[/tex]
