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avatar+1442 

P is the midpoint of line BD.  AP = BP = 4, line AP is perpendicular to line BD, line BD is perpendicular to line DC, line AB is perpendicular to line BC. In simple radical form, what is the perimeter of pentagon ABCDP?

 

 

This one was really tricky for me.  Thanks in advance!

 #1
avatar+88899 
+1

Not too bad, ACG  !!!

 

Note that...if AP  = BP....then triangle APB is an  isosceles right triangle

 

So.....AB  =  sqrt (4^2 + 4^2)  =  sqrt (32)   =  sqrt (16 * 2) =  4sqrt (2)

 

If  P  is the midpoint of BD....then   BP  = 4  and BD  = 8

 

Note that BDC is a right angle....and since angle B is bisected, then  CBD  = 45°  .....so  since BDC  = 90°......then , by default  angle BCD  = 45°

 

So...triangle BDC  is also isosceles with BD  = DC  = 8

 

So......BC  = sqrt  (8^2 + 8^2)  =  sqrt (128)  = sqrt (64* 2)  =  8 sqrt (2)

 

So....the perimeter is

 

BA + AP + PD + DC + BC =

 

4sqrt (2) + 4 + 4 + 8   + 8sqrt (2)  =

 

16  + 12sqrt (2) units  ≈   32.97  units

 

 

cool cool cool

CPhill  Mar 2, 2018
 #2
avatar+1442 
+1

Thank you!  I think I missed something obvious in the lesson.  Do you think you could explain this next question just in case?


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