Atasat este un singur desen pentru ambele puncte a) si b).
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[tex]\displaystyle\bf\\Se~da:\\\triangle ABC cu A = 90^o si D\in BC~(piciorul~inaltimi~duse~din~A)\\\\ Se~cere:\\a)~[AD]~stiind~ca~CB=30~cm~si~2BD=3CD\\b)~[CD]~si~[BD]~stiind~ca~AD=15~cm~si~BD=3CD\\\\Rezolvare:\\a)\\2BD=3CD\\\\ \frac{BD}{3}=\frac{CD}{2}=k\\\\BD=3k\\CD=2k\\BC=BD+CD=3k+2k=30~cm\\3k+2k=30~cm\\5k=30~cm\\\\k=\frac{30}{5}\\\\k=6\\BD=3k=3\times6=18~cm\\CD=2k=2\times6=12~cm\\Aplicam~teorema~inaltimii:\\AD=\sqrt{BD\times CD}=\sqrt{18\times 12}=\sqrt{216}=\boxed{\bf6\sqrt{6}~cm}[/tex]
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[tex]\displaystyle\bf\\b)\\BD=3CD\\\\\frac{BD}{3}=\frac{CD}{1}=k\\\\BD=3k\\CD=k\\Aplicam~teorema~inaltimii:\\\\AD=\sqrt{BD\times CD}=\sqrt{3k\times k}=\sqrt{3k^2}=15~cm\\\\\sqrt{3k^2}=15~~\Big| Ridicam~la~a~2-a\\3k^2=15^2\\3k^2=225\\\\k^2=\frac{225}{3}\\\\k^2=75\\\\ k=\sqrt{75}\\ k=5\sqrt{3}\\BD=3k=3\times5\sqrt{3}=\boxed{\bf15\sqrt{3}}\\CD=k=1\times5\sqrt{3}=\boxed{\bf5\sqrt{3}}[/tex]
