Question

cum se face exercitiul d? mia dat dc= 20 si bc= 900 dar nu cred ca e corect

Answer 1

   

$\displaystyle\bf\\a)\\Se~da:~~BD=4~cm;~~DC=9~cm\\Se~cere:~~BC=?;~~AD=?\\Rezolvare:\\BC=BD+DC=4+9=13~cm\\AD=\sqrt{BD\times DC}=\sqrt{4\times9}=\sqrt{36}=6~cm\\\\b)\\Se~da:~~AD=10~cm;~~BD=5~cm\\Se~cere:~~BC=?;~~DC=?\\Rezolvare:\\AD^2=BD\times DC\\10^2=5\times DC\\DC=\frac{10^2}{5}=\frac{100}{5}=20cm\\BC=BD+DC=5+20=25~cm\\\\c)\\Se~da:~~BC=34~cm;~~DC=25~cm\\Se~cere:~~AD=?;~~BD=?\\Rezolvare:\\BC=BD+DC\\34=BD+25\\BD=34-25=11~cm\\AD=\sqrt{BD\times DC}=\sqrt{11\times 25}=5\sqrt{11}$

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$\displaystyle\bf\\d)\\Se~da:~~AD=30~cm;~~BD=45~cm\\Se~cere:~~BC=?;~~DC=?\\Rezolvare:\\AD^2=BD\times DC\\30^2=45\times DC\\DC=\frac{30^2}{45}=\frac{900}{45}=20~cm\\BC=BD+DC=45+20=65\\\\e)\\Se~da:~~BC=7~cm;~~DC=4~cm\\Se~cere:~~AD=?;~~BD=?\\Rezolvare:\\BD=BC-DC=7-4=3~cm\\\\AD=\sqrt{BD\times DC}=\sqrt{3\times4}=2\sqrt{3}~cm$

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$\displaystyle\bf\\f)\\Se~da:~~AD=12~cm;~~BD=8\sqrt{2}~cm\\Se~cere:~~BC=?~~DC=?\\Rezolvare:\\AD^2=BD\times DC\\ DC=\frac{AD^2}{BD}=\frac{12^2}{8\sqrt{2}}=\frac{144}{8\sqrt{2}}=\frac{144\sqrt{2}}{16}=9\sqrt{2}~cmBC=BD+DC=8\sqrt{2}+9\sqrt{2}=17\sqrt{2}$