Triunghiul ABC este dreptunghic, m(A)=90 grade, AD⊥BC rezulta din teorema inaltimii [tex]AD^2=DB*DC[/tex]
a)BD=BC-DC
DB=70-63=7
[tex]AD^2=DB*DC[/tex]=7*63
AD=[tex] \sqrt{7*63}=21 [/tex]
b)DC=BC-BD=26-8=18
AD=[tex] \sqrt{DB*DC}= \sqrt{18*26}=6 \sqrt{13} [/tex]
c)[tex]AD^2=DB*DC[/tex], deci
[tex]12^2=BD*16[/tex]
BD=144/16=9
BC=BD+DC=9+16=25
d)[tex]AD^2=DB*DC[/tex], deci
[tex]12^2=DC*27[/tex]
BD=144/27=6/3
BC=BD+DC=27+6/3=87/3
e)BC=BD+DC=0.25+5=5.25
[tex]AD^2=DB*DC=0.25*5=1.25[/tex]
[tex]AD=\sqrt{1.25}=0.5\sqrt{5}[/tex]
f)DC=BC=BD=6.5-1.5=5
[tex]AD^2=DB*DC=1.5*5=7.5[/tex]
[tex]AD=\sqrt{7.5}=0.5\sqrt{30}[/tex]
