Question

Buna.Ma ajuta cineva va rog la problema 28

Answer 1

sin60°=AD/AC=>√3/2=12/AC=>AC=8√3
DC²+AD²=AC²=>DC²=AC²-AD²=>DC²=192-144=>DC=4√3
In ΔADB dr. si isoscel [m(ABD)=45°]=>BD=AD=12
sin45°=AD/AB=√2/2=>12/AB=√2/2=>AB=12√2
BC=BD+DC=12+4√3=4(3+√3)

Perim=AB+AC+BC=12√2+8√3+4(3+√3)=12√2+8√3+12+4√3=
=12√2+12√3+12=12(√2+√3+1)