Question 1200556
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The shortest path will be along one of the edges and then diagonally across one of the faces.  You get the most bang for your buck by crossing the largest face on the 
diagonal. So:


Recognizing that *[tex \Large 16\ =\ 4\ \times\ 4] and *[tex \Large 36\ =\ 4\ \times\ 9], we can say *[tex \Large \sqrt{16^2\ +\ 36^2}\ =\ 4\sqrt{4^2\ +\ 9^2}\ =\ 4\sqrt{97}] and then the total shortest distance from A to B is *[tex \Large 4\sqrt{97}\ +\ 12]cm.

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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