Ad[tex] AD^{2} + MD^{2} =
AM^{2}
AD^{2} =
100-36
AD=
\sqrt{64}
[/tex]
AD=8 cm
BC=20 cm
BD=16cm
DC=4cm
În ΔABC:
[tex] AB^{2}=BD*BC
AB= \sqrt{16*20}
AB= \sqrt{320}
AB= 8 \sqrt{5}cm [/tex]
Teorema lui Pitagora:
[tex] AB^{2}+ AC^{2} = BC^{2}
AC= \sqrt{400-320}
AC= \sqrt{80}
AC= 4 \sqrt{5} [/tex]
PERIMETRUL =AB+AC+BC
[tex] 8\sqrt{5} + 4 \sqrt{5} + 20=
12 \sqrt{5}+20=
4(3 \sqrt{5} + 5)[/tex] cm
