a) BD= BC-CD=25-16=9
[tex] AD^{2} = BD X DC= 16 X 9 = 144 \\
AD=12
\\ AB= \sqrt{ AD^{2}+ BD^{2} } = \sqrt{81+144} = \sqrt{225} =15 [/tex] Pitagora in triunghi ABD
[tex] AC^{2} = \sqrt{ AD^{2}+ DC^{2} } = \sqrt{144+256} = \sqrt{400} =20[/tex] Pitagora in triunghi ADC
b) BC=20 BD=16 ; DC=BC-BD=20-16=4
[tex] AD^{2}=BD X BC=16 X 4=64 \\
AD=8 \\
AC^{2} = \sqrt{64 +16} =4 \sqrt{5 } \\
AB^{2} = \sqrt{256 + 64} =8 \sqrt{5} \\ [/tex]
formulele le ai scrise la pct a AC si AB !
