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\(P\) 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 \(ABCDP\)?\(\)

 Mar 16, 2021
 #1
avatar+373 
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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

 Mar 16, 2021
 #2
avatar+15 
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Thanks for explaining, your answer is right :)

alliegoseo5  Mar 17, 2021

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