Răspuns:
a) CD=9,6 cm; BD=5,4 cm; AD=7,2cm; BC=15cm; AB=9cm; AC=12cm
b) BD=14,4cm; DC=25,6cm; AB=24cm; AC=32cm; AD=19,2cm
Explicație pas cu pas:
-----------------------------------------rezolvare a)-----------------------------------------------
[tex]\frac{CD}{BD} =1\frac{7}{9} =\frac{7+9}{9} =\frac{16}{9} \\\\\\\frac{CD}{BD}=\frac{16}{9} = > 9CD=16BD[/tex]
4BD-CD=12 /*4
16BD-4CD=48
Dar 16BD=9CD
deci 9CD-4CD=48
5CD=48
CD=48/5
CD=9,6 cm
4BD-CD=12
4BD=12+9,6=21,6
BD=21,6:4
BD=5,4 cm
Teorema inaltimii:
AD²=BD*CD
[tex]AD=\sqrt{BD*CD} =\sqrt{5,4*9,6} = 7,2\ cm[/tex]
BC=BD+CD
BC=5,4+9,6
BC=15cm
TEOREMA CATETEI:
AB²=BD*BC
[tex]AB=\sqrt{BD*BC} =\sqrt{5,4*15} = 9\ cm[/tex]
TEOREMA CATETEI:
AC²=DC*BC
[tex]AC=\sqrt{CD*BC} =\sqrt{9,6*15} = 12\ cm[/tex]
-----------------------------------------rezolvare b)-----------------------------------------------
[tex]CD-BD=\frac{56}{5} =11,2\\CD=11,2+BD\\\\BC=BD+DC=40\\CD=11,2+BD\\BC=11,2+BD+BD=40\\2BD+11,2=40\\2BD=40-11,2\\2BD=28,8\\BD=28,8:2[/tex]
BD=14,4cm
DC=11,2+BD
DC=11,2+14,4
DC=25,6cm
TEOREMA CATETEI:
AB²=BD*BC
[tex]AB=\sqrt{BD*BC} =\sqrt{14,4*40} = 24\ cm[/tex]
TEOREMA CATETEI:
AC²=DC*BC
[tex]AC=\sqrt{CD*BC} =\sqrt{25,6*40} = 32\ cm[/tex]
Teorema inaltimii:
AD²=BD*CD
[tex]AD=\sqrt{BD*CD} =\sqrt{14,4*25,6} = 19,2\ cm[/tex]
