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!

AnonymousConfusedGuy Mar 2, 2018

#1**+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

CPhill Mar 2, 2018

#2**+1 **

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

AnonymousConfusedGuy
Mar 2, 2018