[tex] h=\frac{l\sqrt{6}}{3}=\frac{6\sqrt{6}}{3}=2\sqrt{6} \ cm\\ [/tex]
< ( (ABC),(DBC) ) = <DMO
[tex] DM=\frac{l\sqrt{3}}{2}=\frac{6\sqrt{3}}{2}=3\sqrt{3} \ cm\\ \sin \angle DMO=\frac{DO}{DM}=\frac{2\sqrt{6}}{3\sqrt{3}}=\bold{\frac{2\sqrt{2}}{3}}[/tex]
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Punctul a)
[tex] h=\frac{l\sqrt{6}}{3}=\frac{6\sqrt{6}}{3}=2\sqrt{6} \ cm\\ [/tex]
Punctul.b)
< ( (ABC),(DBC) ) = <DMO
[tex] DM=\frac{l\sqrt{3}}{2}=\frac{6\sqrt{3}}{2}=3\sqrt{3} \ cm\\ \sin \angle DMO=\frac{DO}{DM}=\frac{2\sqrt{6}}{3\sqrt{3}}=\bold{\frac{2\sqrt{2}}{3}}[/tex]