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Explanation:
Each cube has a side length of 4. Placed together like this, the total horizontal side combines to 4+8 = 8. This is the segment HP as shown in the diagram below. I've also added point Q to form triangle HPQ. This is a right triangle so we can find the hypotenuse QH
Use the pythagorean theorem to find QH
a^2 + b^2 = c^2
(HP)^2 + (PQ)^2 = (QH)^2
8^2 + 4^2 = (QH)^2
(QH)^2 = 64 + 16
(QH)^2 = 80
QH = sqrt(80)
Now we use segment QH to find the length of segment EH. Focus on triangle HQE, which is also a right triangle (right angle at point Q). Use the pythagorean theorem again
a^2 + b^2 = c^2
(QH)^2 + (QE)^2 = (EH)^2
(EH)^2 = (QH)^2 + (QE)^2
(EH)^2 = (sqrt(80))^2 + (4)^2
(EH)^2 = 80 + 16
(EH)^2 = 96
EH = sqrt(96)
EH = sqrt(16*6)
EH = sqrt(16)*sqrt(6)
EH = 4*sqrt(6), showing the answer is choice C
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A shortcut is to use the space diagonal formula. As the name suggests, a space diagonal is one that goes through the solid space (rather than stay entirely on a single face; which you could possibly refer to as a planar diagonal or face diagonal).
The space diagonal formula is
d = sqrt(a^2+b^2+c^2)
which is effectively the 3D version of the pythagorean theorem, or a variant of such.
We have a = HP = 8, b = PQ = 4, and c = QE = 4 which leads to...
d = sqrt(a^2+b^2+c^2)
d = sqrt(8^2+4^2+4^2)
d = sqrt(96)
d = sqrt(16*6)
d = sqrt(16)*sqrt(6)
d = 4*sqrt(6), we get the same answer as before
The space diagonal formula being "pythagorean" in nature isn't a coincidence. Repeated uses of the pythagorean theorem is exactly why this is.
