2. ducem BE_I_ AC AD=CE=18
cos 60⁰ =EB/CE
1/2=EB/18
EB=9
AB=12√3+9
CB=√(9²+18²)=√(81+324)=√405=9√5
P=12√3+18+9√5+12√3+9=27+24√3+9√5
AC=√[(12√3)²+18²]=√(432+324)=√756=6√21
BD=√[(12√3+9)²+18²]=√(432+81+216√3+324)=√(837+216√3)
1.
o sa consider unghiul C=60⁰
tg C=AD/DC
√3=6/DC
DC=2√3
AD²=2√3 xBD
36=2√3 xBD
AC=√[(2√3)²+6²]=√(12+36)=√48=4√3
BD=36/2√3=18√3/3=6√3
BC=2√3+6√3=8√3
AB=√(8√3)²-(4√3)²=√(192-48)=√144=12
P=12+8√3+4√3=12+12√3=12(1+√3)