+15 \(P\) is the midpoint of \(\overline{BD}\) . \(AP\)=\(BP\)=4, \(\overline{AP} \perp \overline{BD}\), \(\overline{BD} \perp \overline{DC}\), \(\overline{AB} \perp \overline{BC}\). 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= \(\sqrt{128}\) = \(8\sqrt{2}\)
BP^2 + AP^2 = AB^2
4^2 + 4^2 = 16 + 16= 2*16=\(\sqrt{32}\) = \(4\sqrt{2}\)
4sqrt2 + 4 + 4 + 8 + 8sqrt2 = 16 + 12sqrt2
