求周界,help me....

(1+1)^2+1^2=AB^2 (1+1)^2+1^2=BC^2
√5=AB √5=BC

(1+1)^2+1^2= AD^2 (1+1)^2+1^2=CD^2
√5=AD √5=CD

AB+BC+AD+CD = 周界

√5+√5+√5+√5 = 4x√5 = 周界

ANS = 4√5

用左pythagrous theorm.

sqrt(2^2+1^2)x4
=4sqrt5

求周界ABCD....SolAB=BC=CD=AD=√(1^2+2^2)=√5周界=4√5

4CM~~~~~~~~~~~~