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# Perimeter Question

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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
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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
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