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April 8th, 2018, 06:11 AM   #1
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[ASK] P to AC

Given an ABCD.EFGH cube whose its side length is 8 cm. The point P is within AB so that AP = 3PH. The distance between P to AC is ....
A. $\displaystyle 2\sqrt3$ cm
B. $\displaystyle 3\sqrt3$ cm
C. $\displaystyle 2\sqrt6$ cm
D. $\displaystyle 3\sqrt6$ cm
E. $\displaystyle 4\sqrt6$ cm

So AH is $\displaystyle 8\sqrt2$ cm and PH = $\displaystyle 6\sqrt2$. What do I do now? Is the triangle HPC a right triangle?
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April 8th, 2018, 06:45 AM   #2
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Something's incorrect in your description.
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April 8th, 2018, 07:02 AM   #3
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Sorry, I mean $\displaystyle PH=2\sqrt2$ cm.
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