+15 P is the midpoint of ¯BD . AP=BP=4, ¯AP⊥¯BD, ¯BD⊥¯DC, ¯AB⊥¯BC. In simplest radical form, what is the perimeter of pentagon ABCDP?
+505 2BP=BD
2*4=8
BD=DC
BD^2 + DC^2 = BC^2
8^2 + 8^2 = 64 + 64 = 2*64= √128 = 8√2
BP^2 + AP^2 = AB^2
4^2 + 4^2 = 16 + 16= 2*16=√32 = 4√2
4sqrt2 + 4 + 4 + 8 + 8sqrt2 = 16 + 12sqrt2
